Understanding Convergence and Divergence in Series

In this video, I'm going to give you a review of the different tests that you need to know in order to determine if a series is going to converge or if it's going to diverge. So the first test we're going to talk about is the divergence test. Now what you need to do is take the limit as n approaches infinity of the sequence a sub n. If this limit does not equal 0, let's say if it equals 5, 1 half infinity. If it doesn't equal 0, then the series diverges. If it equals 0, the series may converge or may diverge. You need to use another test. Now, the next thing we need to talk about is the geometric series. So first, you need to identify the form that it comes in. And it usually looks something like this. There's going to be a constant a, and then your common ratio r is typically raised to the n or n minus 1 power. But your goal is to identify the value of r. So, if r, that is if the absolute value of r is less than 1, then the series is going to converge. Now, if the absolute value of r is greater than or equal to 1, then the series will diverge. Now, the next test you need to be familiar with is the p-series test. So the p-series is in the form of 1 over n raised to the p. So what you need to know is that if p is greater than 1, the series will converge. But if p is less than or equal to 1, then the series is going to diverge. Next, we have the telescoping series. Now, let's say if you have a series where when you write out the terms, it's like... 1 minus 1 over 2 plus 1 over 2 minus 1 over 3 plus 1 over 3 minus 1 over 4. And typically, you can cancel the terms in the middle. Now what you want to do, because this will go to the general formula a sub n, you want to write a formula that will give you the partial sum of the telescope in series. It could be 1 minus... a sub n or 1 plus a sub n or something of that nature. But once you write the general formula for the partial sum, then you want to evaluate the sum of the series as n approaches infinity, which is going to be the limit as n goes to infinity of the partial sum formula. Now, if you get some finite value, let's call L, then that means that the series converges. which is usually the case. Now if you get, let's say, plus or minus infinity or it doesn't exist, then the series is divergent. So here is a typical example of a telescoping series problem. Let's say if you have the series from 1 to infinity of 1 over n minus 1 over n plus 1. That's a telescoping series problem. If you list out the terms, it's going to be 1 over 1 minus 1 over 2, and then plus 1 over 2 minus 1 over 3, and so forth. Now sometimes, you may not have two separate fractions. You may only have a single fraction. And you need to use partial fraction decomposition to split. split it into two separate fractions and then you can list out the terms. So watch out for the telescope in series problem. You need to keep an eye out for that one. Now the next test you need to be familiar with is the integral test. So what do you recall about the integral test? So let's say if we have the sequence a sub n, and we're going to make it equal to a function of n. f has to be positive, continuous, and a decrease in function. And it has to be decreasing from some number n to infinity. So the long-term trend is it has to be decreasing. Sometimes it can go up initially, but as long as it's decreasing for the long run, it can work. So what you need to do is take the integral from 1 to infinity of f of x dx, and let's say if you get some finite value out, then that means that the series, the original series, converges. Now let's say if you get plus or minus infinity or you get an answer where it doesn't exist then the original series diverges and so that's the integral test. The next one we're going to talk about is the ratio test. In this case, you need to take the limit as n goes to infinity of the absolute value of a sub n plus 1 divided by a sub n. Now, if your result is less than 1, then the series converges. If your result is greater than 1, or if it goes to, let's say, plus positive infinity, negative infinity won't apply here because your final answer has to be positive. Then the series will diverge. Now, if the limit is equal to 1, Then the ratio test is inconclusive. Next up is the root test. So what you need to do is take the limit as n goes to infinity of the f root of the absolute value of a sub n. Now, if you get an answer that's less than 1, then the series converges. Now, if you get a value where it's greater than 1, or if it goes to infinity, then the series converges. diverges. If your answer is equal to one, then like the ratio test, the root test will become inconclusive and so you may need to use another test to get the answer to determine if the series is going to converge or diverge. Now let's talk about the direct comparison test. What do you remember about this test? Well first, we need to have two sequences, a sub n and b sub n. So let's say b sub n is the big sequence and a sub n is the small sequence. And so here's the basic idea behind the direct comparison test. So if the big series converges, then the small series will converge as well. Likewise, if the small series diverges, then the big series diverges as well. Now, let's talk about the limit comparison test. This test is very similar to the ratio test. The only difference is you're dealing with two different sequences as opposed to one. So you need to take the limit as n goes to infinity of a sub n divided by b sub n. You need to show that this is equal to some finite positive number which we'll call L and if that's the case then both series will either converge or both series will diverge. So let's say if you know that. this series converges and if you pass the limit comparison test then you can say that the series with B sub n that's going to converge as well or let's say if this series diverges then you know this series the other one will diverge as well So either they both converge or they both diverge. And that's the basic idea behind the limit comparison test. If you know the convergence or divergence of one series, then you can determine the convergence or divergence of another series. Next up is the alternating series test. And this is when you have a series with alternating signs. So typically, it's in this form. You're going to have negative 1 raised to the n-or n plus 1 times some sequence a sub n. And in order to determine if the series will converge, two conditions must be met. The first condition is that it has to pass the divergence test. So you need to take the limit as n goes to infinity, and you need to show that it's 0. If it's not 0, then the series diverges. If it's 0, then you need to move on to step 2. You need to show that the sequence is decreasing. So a sub n has to be equal to or greater than a sub n plus 1. The next thing you could try is the absolute value test. So let's say if the absolute value of the series converges, then the original series will also converge. And when this happens you can say that the series, the original series, is absolutely convergent. Now, let's say if you take the absolute value of the series and you find that it diverges. And you analyze the original series, but it doesn't diverge. Let's say it's convergent. If the original series is convergent, then the original series is conditionally convergent. Now. Let's say if you analyze the absolute value of the series, and you find it to be divergent. Then you check the original series, and that's divergent as well. Then the whole series is simply divergent. There's nothing to be said about it. So let's say if we have the series from 1 to infinity, 2n squared plus 5. over 7n squared minus 4. So what do you think we need to do for this problem? Which test should we use? If you're not sure, it's always good to start with the divergence test. So let's take the limit as n goes to infinity of a sub n. Now, a sub n is what we see here. So what's the limit equal to? If you want to, you can use L'Hopital's rule. So let's take the derivative of the numerator and the denominator. So the derivative of 2n squared, that's going to be 4n. And the derivative of 7n squared is 14n. And then what we could do is use L'Hopital's rule one more time. And so this is going to be 4 over 14, which reduces to 2 over 7. So notice that the limit, as n goes to infinity of a sub n, it doesn't equal 0. Therefore, the original series is divergent, according to the divergence test. And that's it for this problem. Consider this series. Let's say if we have the cube root of n divided by n to the fifth power. Will the series converge or will it diverge? So which test would you use here? Notice that we can put the series in this form. You can simplify it or reduce it to this form. So we're dealing with a p-series. So the cube root of n, that's n to the 1 third, and let's divide it by n to the 5th. So we need to subtract the exponents. What's 1 over 3 minus 5? So again, in common denominators, let's multiply this by 3 over 3. And so this is 1 over 3 minus 15 over 3, so that's going to be negative 14 over 3. So we can rewrite the series like this. So we have n to the negative 14 over 3, which we can rewrite that as 1 over n to the 14 over 3. So now it's in the proper format. So we can clearly see that p is equal to 14 over 3. And 14 over 3 is greater than 1, so p is greater than 1. And for the p-series, if p is greater than 1, then the series converges. Now let's try another problem. So let's say this is 5 times 1 fourth raised to the n minus 1. What type of series do we have? Notice that it's in this form. So what we have is a geometric series. So the common ratio is 1 over 4. Now recall that if the absolute value of r is less than 1, the series converges for geometric series. And if it's greater than or equal to 1, the series diverges. 1 fourth is certainly less than 1, so we can say that this particular geometric series, it converges. Now what if we wanted to find the sum of the geometric series as it goes to infinity? If r is less than 1, then the sum is equal to the first term divided by 1 minus the common ratio. So to find the first term, we need to plug in n equal 1. So that's going to be 5 times 1 over 4 raised to the 1 minus 1, or 0. 1 fourth raised to the 0 power is 1, and so we get 5. So 5 is the... the first term. So it's going to be 5 divided by 1 minus 1 fourth. So we need to multiply the top and the bottom by 4. 5 times 4 is 20, and then distribute the 4. 4 times 1 is 4, and 4 times the fourth is 1. So the sum is going to be 20 over 3. So the value of this series is equal to this number. Now let's try this one. So let's say we have negative 1 raised to the n power divided by the square root of n. So what type of series are we dealing with here? Well, let's write out the first few terms. When n is 1, it's going to be negative 1 divided by the square root of 1 and when n is 2 is going to be negative 1 squared is positive 1 divided by the square root of 2 and then it's gonna be negative 1 over the square root of 3 and so forth so notice that we have an alternating series So what should we do here? How can we determine if the alternating series will converge or diverge? Two conditions must be met. First, it has to pass the divergence test. So if we take the limit as n approaches infinity of a sub n. Now you need to know what a sub n is in this problem. So for the alternating series, it's typically in this form. It's going to be negative 1 to the n power times a sub n. So a sub n is everything except this. This could be n plus 1 too, by the way. Just keep that in mind. So a sub n is 1 divided by the square root of n. So if we take the limit as n goes to infinity of 1 divided by the square root of n, then this is going to go to 0. So it passes the divergence test. Now the second thing that we need to show is that the next term is less than or equal to the previous term. So the previous term being 1 divided by the square root of n, and the next term is going to be 1 over the square root of n plus 1. So the denominator of this fraction is bigger, which means the value of this fraction is smaller. So this is a true statement. Therefore, we could say... that this series, according to the alternating series test, is convergent. Now, is it absolutely convergent, or is it conditionally convergent? What would you say? So let's analyze the convergence of the absolute value of the series. So, if we take the absolute value of negative 1 to the n, divided by the square root of n, the negative 1 to the n, it produces the alternating signs. 1, negative 1, negative 1, 1, and so forth. So, if you take the absolute value of that, the negative 1 and the 1 will just be 1. So, then we're going to get this series. So, it's going to be 1 divided by the square root of n. And the square root of n is n to the 1 half. So, what type of series do we have? Notice that we have the p-series. It's in the form 1 over n to the p. So we can clearly see that p is 1 half. So if p is 1 half, will the series converge or will it diverge? In order for the series to converge, p has to be greater than 1. But because p is less than or equal to 1, this series is divergent. So because the original series... converges, but the absolute value of the series diverges, then we say that the original series is conditionally convergent. In order for it to be absolutely convergent, the original series must converge, and the absolute value of the series must converge as well. Now let's move on to our next example. So let's say this is 1 over n times n plus 1. Will the series converge or diverge? Well, let's check with divergence tests. So what happens when n goes to infinity? So we can rewrite this as the limit as n goes to infinity of 1 over n squared plus n. And if we multiply the top and the bottom by 1 over n squared, we're going to get the limit as n goes to infinity, 1 over n squared divided by 1 plus 1 over n. So 1 over n squared, that's going to go to 0. And 1 over n is going to go to 0. So it equals 0. So the divergence test is inconclusive. So what do you think we need to do here? This problem can be solved if we recognize it to be a telescope in series problem. So first, we need to use partial fraction decomposition. So let's set 1 over n times n plus 1 to a over n plus b over n plus 1. And let's multiply both sides by n times n plus 1. So these will cancel, and that's going to give us 1. And then n will cancel, leaving behind 8. times n plus 1 and then n plus 1 will cancel leaving behind b times n. Now let's say if n is equal to 0 that means bn disappears so 1 is equal to a times 0 plus 1, which means a is equal to 1. Now, if n is equal to negative 1, this will disappear. That will be 0. So, 1 is equal to b times negative 1, which means b is negative 1. So we can rewrite this expression like this. So instead of 1 over n times n plus 1, we can replace a with 1, so it's going to be 1 over n. And then we can replace b with negative 1, so it's negative 1 over n plus 1. So now you can see that we have a telescope in series. So let's write out a few terms. So when n is 1, this is going to be 1 over 1 minus 1 over 2. So this is the first term. And then the second term, when n is 2, is going to be plus 1 over 2 minus 1 over 3. So that's a sub 2 and then when n is 3 is going to be 1 over 3 minus 1 over 4 So that's the third term Now let's write the a sub n minus 1 term. So if we replace this with n minus 1, it's just going to be 1 over n minus 1. And then if we replace this with n minus 1 plus 1, it's just going to be... 1 over n. Now the last term, that is the a sub n term, is just going to be what you see here. So 1 over n minus 1 over n plus 1. So all of this is the sum of the first n terms. So this is the partial sum. Now let's see what we can cancel. So we can cancel negative 1 half and 1 half, negative 1 third and 1 third. And then negative 1 fourth will cancel with 1 fourth, which is somewhere in the middle of the sequence. And then we can cancel. negative 1 over n plus 1 over n. Now 1 over n minus 1 will cancel with another 1 over n minus 1 if you write out the a sub n minus 2 term. So the only thing that's going to remain is 1 and negative 1 over n plus 1. So this allows us to write the general formula for the partial sum. So it's 1 minus 1 over n plus 1. Let's get rid of this too. So now we need to calculate the sum of the sequence. So let's take the limit as n approaches infinity of the partial sum formula S sub n. So that's 1 minus 1 over n plus 1. So this is equal to 1 minus the limit as n goes to infinity of 1 over n plus 1. And this is going to go to 0. So it's 1 minus 0, which is 1. So therefore, the sum of this series is equal to 1. And because it's equal to a finite positive number and not infinity, the original series, we say, converges. Here's another one that you could try. Let's say if it's 1 over n squared plus 4. So what can we do for this problem? How can we determine if the series is going to converge or diverge? What test would you recommend? The best thing to do is to compare it, to use the direct comparison test. Notice that if we take away the 4, this becomes a p-series problem. Now, which sequence is bigger? 1 over n squared plus 4, or 1 over n squared? 1 over n squared is greater than 1 over n squared plus 4. If you plug in 1, this would be 1, and this is going to be 1 over 5. So this is always going to be bigger for all n. So therefore, we could say this is our a sub n term, and this is our b sub n term. So this series is bigger than this one. Now, according to the p-series, we see that p is equal to 2. We know this series converges. Because p is greater than 1. Now, according to the direct comparison test, if the big series converges, then the small series will converge as well. And that's all we need to do for this problem. So this is going to be convergent. Now let's move on to our next problem. So let's say if we have a series which starts from 3 and goes to infinity, and it's 1 over the square root of n minus 2. So what sort of tests should we use in order to determine the convergence or the divergence of the series? Well, we can start with the divergence test. But when n becomes very large, because it's bottom heavy, the limit will go to 0. And so, based on a divergence test, it may diverge or may converge. Next, we could try the direct comparison test, because if we get rid of the 2, then this becomes a p-series. And the square root of n is n to the 1 half, which means that p is 1 half. So because p is less than or equal to 1, this series diverges. Now, which series is bigger? Is it this one or this one? Because n minus 2 is less than n, we have a smaller denominator. If you decrease the value of the denominator of a fraction, the value of the whole fraction goes down. I mean, excuse me, the value of the whole fraction goes up. So therefore, this series is larger, so we should write greater than or equal to. So this is going to be the big series, b sub n, and this is going to be the small series, a sub n. So, if the small series diverges, will the big series diverge as well? And the answer is yes. The big series must diverge as well. So, this could work. Now, there's another test that you could use if you want to. And that is the integral test. So let's say f of x is 1 divided by the square root of x minus 2. Now the function is not continuous at x equals 2, because that's a vertical asymptote. But we're starting from 3. So it is continuous from 3 to infinity. We need to show that it's continuous, decreasing, and positive. This is always going to be positive. especially when x is greater than 3. And because we have a square root of x minus 2 on the bottom, it's going to be a decrease in function. As x increases, the value of the denominator will go up, and so the function will decrease. Now, you can also find the first derivative and show that it's negative on this interval. If you do that, that tells you that the function is decreasing. So now, once those conditions are met, once you show that it's positive, continuous, and decreasing on this interval, take the integral, in this case, starting from 3 to infinity, sometimes it's usually 1 to infinity, integrate f of x dx from a to b. So what we have here is an improper integral. So first, I'm going to find the indefinite integral. I'm going to use u substitution. So if I make u equal to x minus 2, du is going to be equal to dx. And so I can replace x minus 2 with u, and dx with du. So I can rewrite this as u to negative 1 half, and so that becomes u to the positive 1 half, divided by 1 half, or multiplied by 2. And so this becomes 2 times the square root of x minus 2. So this is equal to the limit as a goes to infinity of 2 times the square root of x minus 2 evaluated from 3 to a. So now, let's plug in a and 3. So we're going to have the limit as a goes into infinity, and this is going to be 2 times the square root of a minus 2, plus 2 times, I mean not plus, but minus, minus 2 times the square root of 3 minus 2. Now, as a goes to infinity, this whole expression becomes infinity. And so infinity minus 2 times 1 is just infinity. So if the integral test gives you infinity, that means that the original series diverges, which was in harmony with the other conclusion that we got based on the direct comparison test. Let's say if we have the series of the square root of n divided by n cubed plus 2. Which test should we use for this one? The test that I recommend using is the limit comparison test. But we need two separate functions, one of which is... more simplified than the other. So for the other series, I'm just going to get rid of the two. So it becomes the square root of n divided by n cubed. And according to the limit comparison test, if we take the limit as n goes to infinity of the ratio of the two sequences, if it's equal to a finite positive number, then both series will either converge or they will both diverge. Now we can reduce this to a p-series. So this is n to the 1 half divided by n cubed and so a half minus 3 is 2.5, negative 2.5. And so this becomes 1 over n to the 2.5. So p is 2.5 and that's greater than 1. So according to the p-series test, this series converges. So if this passes the limit comparison test, it will also converge. Let's find out. Now let's take the limit as n goes to infinity of a sub n divided by b sub n. So let's call this our a sub n. And this expression is going to be our b sub n sequence. So this is going to be the limit as n goes to infinity, and then a sub n, that's the square root of n, divided by n to the third plus 2, and then divided by b sub n, which is the square root of n over n cubed. Perhaps you heard of the expression keep change flip. Keep the first fraction the same, change division to multiplication, flip the second fraction. So we can rewrite this as times n to the third divided by the square root of n. So these two will cancel, and so we have the limit as n goes to infinity. of n cubed divided by n cubed plus 2. So let's multiply the top and the bottom by 1 over n cubed. So if we do that, we're going to get the limit as n approaches infinity, and this is going to be 1 divided by 1 plus 2 times 1 over n cubed. So when n becomes very large, 1 over n cubed goes to 0. So it's 1 over 1 plus 2 times 0, which is 1 over 1, and that's 1. So this is equal to a finite positive number. Therefore, according to the limit comparison test, because this series converges, this one must converge as well. Now, consider the next example. Let's say it's 3n squared minus 9 divided by 7n squared plus 4 raised to the n. Which tests should we use? If you see something raised to the n power, it's a good indication that we need to use the root test. So recall that the root test is this. Take the limit as n goes to infinity of the nth root of the absolute value of the sequence a sub n. Now, if it's less than 1, the series converges. If it's greater than 1 or if it's infinity, the series diverges. If it's equal to 1, then it's inconclusive. So if we take the limit as n goes into infinity of the nth root, which is the same as raising everything to the 1 over n power. n times 1 over n, those will cancel. And so we have the limit as n goes to infinity of 3n squared minus 9 divided by 7n squared plus 4. Notice that the degree of the numerator is the same as that of the denominator. So you can see that the limit is going to be 3 over 7. But if you want to show your work, you can multiply the top and the bottom by 1 over n squared, or use L'Hopital's rule. Let's use L'Hopital's rule for this example. So the derivative of 3n squared is 6n, and the derivative of 7n squared is 14n. So using L'Hopital's rule again. This is going to become 6 over 14, which reduces to 3 over 7. So this is the answer. Now, 3 over 7 is less than 1. So based on the root test, the series converges. Now, let's work on our final example. And so we're going to have this series of 2 raised to the n divided by n factorial. So what tests should we use for this example? So since you know it's the last problem, it's going to be the test that we haven't used yet, which is the ratio test. And so the basic idea behind a ratio test is we need to take... the a sub n plus 1, and divided by a sub n, all within an absolute value symbol. And if this is less than 1, the series converges. If it's greater than 1 or equals infinity, the series diverges. Now, if it's equal to 1, then it's inconclusive. Now, we know that a sub n is 2 to the n over n factorial. But what's a sub n plus 1? All you need to do is replace n with n plus 1. So the limit as n approaches infinity is going to be a sub n and then divided by, I mean, I'll take that back. It's a sub n plus 1. We need to write that one first. And then divided by a sub n. Using keep change flip, we could say it's multiplied by 1 over a sub n, or the reciprocal of a sub n. So this is a sub n plus 1 times 1 over a sub n. So you can represent it that way if you want to. So how can we simplify the limit now? What do we need to do? 2 to the n plus 1, that's 2 to the n times 2 to the first power. n plus 1 factorial is n plus 1 times n factorial. And so we can cancel 2 to the n, and we can cancel n factorial. So we're left with the limit as n goes to infinity of 2 divided by n plus 1. 2 over n plus 1 is going to go to 0 as n goes to infinity. So 0 is less than 1, which means that the series converges by the ratio test. And that's the answer.

If it equals 0, the series may converge or may diverge. You need to use another test. Now, the next thing we need to talk about is the geometric series. So first, you need to identify the form that it comes in.

And it usually looks something like this. There's going to be a constant a, and then your common ratio r is typically raised to the n or n minus 1 power. But your goal is to identify the value of r. So, if r, that is if the absolute value of r is less than 1, then the series is going to converge.

Now, if the absolute value of r is greater than or equal to 1, then the series will diverge. Now, the next test you need to be familiar with is the p-series test. So the p-series is in the form of 1 over n raised to the p. So what you need to know is that if p is greater than 1, the series will converge.

But if p is less than or equal to 1, then the series is going to diverge. Next, we have the telescoping series. Now, let's say if you have a series where when you write out the terms, it's like... 1 minus 1 over 2 plus 1 over 2 minus 1 over 3 plus 1 over 3 minus 1 over 4. And typically, you can cancel the terms in the middle.

Now what you want to do, because this will go to the general formula a sub n, you want to write a formula that will give you the partial sum of the telescope in series. It could be 1 minus... a sub n or 1 plus a sub n or something of that nature. But once you write the general formula for the partial sum, then you want to evaluate the sum of the series as n approaches infinity, which is going to be the limit as n goes to infinity of the partial sum formula. Now, if you get some finite value, let's call L, then that means that the series converges.

which is usually the case. Now if you get, let's say, plus or minus infinity or it doesn't exist, then the series is divergent. So here is a typical example of a telescoping series problem. Let's say if you have the series from 1 to infinity of 1 over n minus 1 over n plus 1. That's a telescoping series problem. If you list out the terms, it's going to be 1 over 1 minus 1 over 2, and then plus 1 over 2 minus 1 over 3, and so forth.

Now sometimes, you may not have two separate fractions. You may only have a single fraction. And you need to use partial fraction decomposition to split.

split it into two separate fractions and then you can list out the terms. So watch out for the telescope in series problem. You need to keep an eye out for that one.

Now the next test you need to be familiar with is the integral test. So what do you recall about the integral test? So let's say if we have the sequence a sub n, and we're going to make it equal to a function of n. f has to be positive, continuous, and a decrease in function.

And it has to be decreasing from some number n to infinity. So the long-term trend is it has to be decreasing. Sometimes it can go up initially, but as long as it's decreasing for the long run, it can work.

So what you need to do is take the integral from 1 to infinity of f of x dx, and let's say if you get some finite value out, then that means that the series, the original series, converges. Now let's say if you get plus or minus infinity or you get an answer where it doesn't exist then the original series diverges and so that's the integral test. The next one we're going to talk about is the ratio test.

In this case, you need to take the limit as n goes to infinity of the absolute value of a sub n plus 1 divided by a sub n. Now, if your result is less than 1, then the series converges. If your result is greater than 1, or if it goes to, let's say, plus positive infinity, negative infinity won't apply here because your final answer has to be positive.

Then the series will diverge. Now, if the limit is equal to 1, Then the ratio test is inconclusive. Next up is the root test. So what you need to do is take the limit as n goes to infinity of the f root of the absolute value of a sub n.

Now, if you get an answer that's less than 1, then the series converges. Now, if you get a value where it's greater than 1, or if it goes to infinity, then the series converges. diverges. If your answer is equal to one, then like the ratio test, the root test will become inconclusive and so you may need to use another test to get the answer to determine if the series is going to converge or diverge. Now let's talk about the direct comparison test.

What do you remember about this test? Well first, we need to have two sequences, a sub n and b sub n. So let's say b sub n is the big sequence and a sub n is the small sequence.

And so here's the basic idea behind the direct comparison test. So if the big series converges, then the small series will converge as well. Likewise, if the small series diverges, then the big series diverges as well. Now, let's talk about the limit comparison test.

This test is very similar to the ratio test. The only difference is you're dealing with two different sequences as opposed to one. So you need to take the limit as n goes to infinity of a sub n divided by b sub n.

You need to show that this is equal to some finite positive number which we'll call L and if that's the case then both series will either converge or both series will diverge. So let's say if you know that. this series converges and if you pass the limit comparison test then you can say that the series with B sub n that's going to converge as well or let's say if this series diverges then you know this series the other one will diverge as well So either they both converge or they both diverge.

And that's the basic idea behind the limit comparison test. If you know the convergence or divergence of one series, then you can determine the convergence or divergence of another series. Next up is the alternating series test.

And this is when you have a series with alternating signs. So typically, it's in this form. You're going to have negative 1 raised to the n-or n plus 1 times some sequence a sub n.

And in order to determine if the series will converge, two conditions must be met. The first condition is that it has to pass the divergence test. So you need to take the limit as n goes to infinity, and you need to show that it's 0. If it's not 0, then the series diverges. If it's 0, then you need to move on to step 2. You need to show that the sequence is decreasing.

So a sub n has to be equal to or greater than a sub n plus 1. The next thing you could try is the absolute value test. So let's say if the absolute value of the series converges, then the original series will also converge. And when this happens you can say that the series, the original series, is absolutely convergent.

Now, let's say if you take the absolute value of the series and you find that it diverges. And you analyze the original series, but it doesn't diverge. Let's say it's convergent. If the original series is convergent, then the original series is conditionally convergent.

Now. Let's say if you analyze the absolute value of the series, and you find it to be divergent. Then you check the original series, and that's divergent as well. Then the whole series is simply divergent. There's nothing to be said about it.

So let's say if we have the series from 1 to infinity, 2n squared plus 5. over 7n squared minus 4. So what do you think we need to do for this problem? Which test should we use? If you're not sure, it's always good to start with the divergence test. So let's take the limit as n goes to infinity of a sub n.

Now, a sub n is what we see here. So what's the limit equal to? If you want to, you can use L'Hopital's rule. So let's take the derivative of the numerator and the denominator.

So the derivative of 2n squared, that's going to be 4n. And the derivative of 7n squared is 14n. And then what we could do is use L'Hopital's rule one more time.

And so this is going to be 4 over 14, which reduces to 2 over 7. So notice that the limit, as n goes to infinity of a sub n, it doesn't equal 0. Therefore, the original series is divergent, according to the divergence test. And that's it for this problem. Consider this series. Let's say if we have the cube root of n divided by n to the fifth power. Will the series converge or will it diverge?

So which test would you use here? Notice that we can put the series in this form. You can simplify it or reduce it to this form.

So we're dealing with a p-series. So the cube root of n, that's n to the 1 third, and let's divide it by n to the 5th. So we need to subtract the exponents.

What's 1 over 3 minus 5? So again, in common denominators, let's multiply this by 3 over 3. And so this is 1 over 3 minus 15 over 3, so that's going to be negative 14 over 3. So we can rewrite the series like this. So we have n to the negative 14 over 3, which we can rewrite that as 1 over n to the 14 over 3. So now it's in the proper format. So we can clearly see that p is equal to 14 over 3. And 14 over 3 is greater than 1, so p is greater than 1. And for the p-series, if p is greater than 1, then the series converges. Now let's try another problem.

So let's say this is 5 times 1 fourth raised to the n minus 1. What type of series do we have? Notice that it's in this form. So what we have is a geometric series.

So the common ratio is 1 over 4. Now recall that if the absolute value of r is less than 1, the series converges for geometric series. And if it's greater than or equal to 1, the series diverges. 1 fourth is certainly less than 1, so we can say that this particular geometric series, it converges.

Now what if we wanted to find the sum of the geometric series as it goes to infinity? If r is less than 1, then the sum is equal to the first term divided by 1 minus the common ratio. So to find the first term, we need to plug in n equal 1. So that's going to be 5 times 1 over 4 raised to the 1 minus 1, or 0. 1 fourth raised to the 0 power is 1, and so we get 5. So 5 is the... the first term.

So it's going to be 5 divided by 1 minus 1 fourth. So we need to multiply the top and the bottom by 4. 5 times 4 is 20, and then distribute the 4. 4 times 1 is 4, and 4 times the fourth is 1. So the sum is going to be 20 over 3. So the value of this series is equal to this number. Now let's try this one. So let's say we have negative 1 raised to the n power divided by the square root of n. So what type of series are we dealing with here?

Well, let's write out the first few terms. When n is 1, it's going to be negative 1 divided by the square root of 1 and when n is 2 is going to be negative 1 squared is positive 1 divided by the square root of 2 and then it's gonna be negative 1 over the square root of 3 and so forth so notice that we have an alternating series So what should we do here? How can we determine if the alternating series will converge or diverge? Two conditions must be met. First, it has to pass the divergence test.

So if we take the limit as n approaches infinity of a sub n. Now you need to know what a sub n is in this problem. So for the alternating series, it's typically in this form. It's going to be negative 1 to the n power times a sub n.

So a sub n is everything except this. This could be n plus 1 too, by the way. Just keep that in mind.

So a sub n is 1 divided by the square root of n. So if we take the limit as n goes to infinity of 1 divided by the square root of n, then this is going to go to 0. So it passes the divergence test. Now the second thing that we need to show is that the next term is less than or equal to the previous term.

So the previous term being 1 divided by the square root of n, and the next term is going to be 1 over the square root of n plus 1. So the denominator of this fraction is bigger, which means the value of this fraction is smaller. So this is a true statement. Therefore, we could say...

that this series, according to the alternating series test, is convergent. Now, is it absolutely convergent, or is it conditionally convergent? What would you say?

So let's analyze the convergence of the absolute value of the series. So, if we take the absolute value of negative 1 to the n, divided by the square root of n, the negative 1 to the n, it produces the alternating signs. 1, negative 1, negative 1, 1, and so forth. So, if you take the absolute value of that, the negative 1 and the 1 will just be 1. So, then we're going to get this series.

So, it's going to be 1 divided by the square root of n. And the square root of n is n to the 1 half. So, what type of series do we have? Notice that we have the p-series. It's in the form 1 over n to the p.

So we can clearly see that p is 1 half. So if p is 1 half, will the series converge or will it diverge? In order for the series to converge, p has to be greater than 1. But because p is less than or equal to 1, this series is divergent. So because the original series...

converges, but the absolute value of the series diverges, then we say that the original series is conditionally convergent. In order for it to be absolutely convergent, the original series must converge, and the absolute value of the series must converge as well. Now let's move on to our next example.

So let's say this is 1 over n times n plus 1. Will the series converge or diverge? Well, let's check with divergence tests. So what happens when n goes to infinity?

So we can rewrite this as the limit as n goes to infinity of 1 over n squared plus n. And if we multiply the top and the bottom by 1 over n squared, we're going to get the limit as n goes to infinity, 1 over n squared divided by 1 plus 1 over n. So 1 over n squared, that's going to go to 0. And 1 over n is going to go to 0. So it equals 0. So the divergence test is inconclusive.

So what do you think we need to do here? This problem can be solved if we recognize it to be a telescope in series problem. So first, we need to use partial fraction decomposition. So let's set 1 over n times n plus 1 to a over n plus b over n plus 1. And let's multiply both sides by n times n plus 1. So these will cancel, and that's going to give us 1. And then n will cancel, leaving behind 8. times n plus 1 and then n plus 1 will cancel leaving behind b times n.

Now let's say if n is equal to 0 that means bn disappears so 1 is equal to a times 0 plus 1, which means a is equal to 1. Now, if n is equal to negative 1, this will disappear. That will be 0. So, 1 is equal to b times negative 1, which means b is negative 1. So we can rewrite this expression like this. So instead of 1 over n times n plus 1, we can replace a with 1, so it's going to be 1 over n.

And then we can replace b with negative 1, so it's negative 1 over n plus 1. So now you can see that we have a telescope in series. So let's write out a few terms. So when n is 1, this is going to be 1 over 1 minus 1 over 2. So this is the first term. And then the second term, when n is 2, is going to be plus 1 over 2 minus 1 over 3. So that's a sub 2 and then when n is 3 is going to be 1 over 3 minus 1 over 4 So that's the third term Now let's write the a sub n minus 1 term.

So if we replace this with n minus 1, it's just going to be 1 over n minus 1. And then if we replace this with n minus 1 plus 1, it's just going to be... 1 over n. Now the last term, that is the a sub n term, is just going to be what you see here. So 1 over n minus 1 over n plus 1. So all of this is the sum of the first n terms. So this is the partial sum.

Now let's see what we can cancel. So we can cancel negative 1 half and 1 half, negative 1 third and 1 third. And then negative 1 fourth will cancel with 1 fourth, which is somewhere in the middle of the sequence.

And then we can cancel. negative 1 over n plus 1 over n. Now 1 over n minus 1 will cancel with another 1 over n minus 1 if you write out the a sub n minus 2 term.

So the only thing that's going to remain is 1 and negative 1 over n plus 1. So this allows us to write the general formula for the partial sum. So it's 1 minus 1 over n plus 1. Let's get rid of this too. So now we need to calculate the sum of the sequence.

So let's take the limit as n approaches infinity of the partial sum formula S sub n. So that's 1 minus 1 over n plus 1. So this is equal to 1 minus the limit as n goes to infinity of 1 over n plus 1. And this is going to go to 0. So it's 1 minus 0, which is 1. So therefore, the sum of this series is equal to 1. And because it's equal to a finite positive number and not infinity, the original series, we say, converges. Here's another one that you could try. Let's say if it's 1 over n squared plus 4. So what can we do for this problem? How can we determine if the series is going to converge or diverge?

What test would you recommend? The best thing to do is to compare it, to use the direct comparison test. Notice that if we take away the 4, this becomes a p-series problem.

Now, which sequence is bigger? 1 over n squared plus 4, or 1 over n squared? 1 over n squared is greater than 1 over n squared plus 4. If you plug in 1, this would be 1, and this is going to be 1 over 5. So this is always going to be bigger for all n.

So therefore, we could say this is our a sub n term, and this is our b sub n term. So this series is bigger than this one. Now, according to the p-series, we see that p is equal to 2. We know this series converges. Because p is greater than 1. Now, according to the direct comparison test, if the big series converges, then the small series will converge as well. And that's all we need to do for this problem.

So this is going to be convergent. Now let's move on to our next problem. So let's say if we have a series which starts from 3 and goes to infinity, and it's 1 over the square root of n minus 2. So what sort of tests should we use in order to determine the convergence or the divergence of the series? Well, we can start with the divergence test. But when n becomes very large, because it's bottom heavy, the limit will go to 0. And so, based on a divergence test, it may diverge or may converge.

Next, we could try the direct comparison test, because if we get rid of the 2, then this becomes a p-series. And the square root of n is n to the 1 half, which means that p is 1 half. So because p is less than or equal to 1, this series diverges.

Now, which series is bigger? Is it this one or this one? Because n minus 2 is less than n, we have a smaller denominator. If you decrease the value of the denominator of a fraction, the value of the whole fraction goes down. I mean, excuse me, the value of the whole fraction goes up.

So therefore, this series is larger, so we should write greater than or equal to. So this is going to be the big series, b sub n, and this is going to be the small series, a sub n. So, if the small series diverges, will the big series diverge as well?

And the answer is yes. The big series must diverge as well. So, this could work. Now, there's another test that you could use if you want to.

And that is the integral test. So let's say f of x is 1 divided by the square root of x minus 2. Now the function is not continuous at x equals 2, because that's a vertical asymptote. But we're starting from 3. So it is continuous from 3 to infinity. We need to show that it's continuous, decreasing, and positive.

This is always going to be positive. especially when x is greater than 3. And because we have a square root of x minus 2 on the bottom, it's going to be a decrease in function. As x increases, the value of the denominator will go up, and so the function will decrease.

Now, you can also find the first derivative and show that it's negative on this interval. If you do that, that tells you that the function is decreasing. So now, once those conditions are met, once you show that it's positive, continuous, and decreasing on this interval, take the integral, in this case, starting from 3 to infinity, sometimes it's usually 1 to infinity, integrate f of x dx from a to b. So what we have here is an improper integral. So first, I'm going to find the indefinite integral.

I'm going to use u substitution. So if I make u equal to x minus 2, du is going to be equal to dx. And so I can replace x minus 2 with u, and dx with du. So I can rewrite this as u to negative 1 half, and so that becomes u to the positive 1 half, divided by 1 half, or multiplied by 2. And so this becomes 2 times the square root of x minus 2. So this is equal to the limit as a goes to infinity of 2 times the square root of x minus 2 evaluated from 3 to a. So now, let's plug in a and 3. So we're going to have the limit as a goes into infinity, and this is going to be 2 times the square root of a minus 2, plus 2 times, I mean not plus, but minus, minus 2 times the square root of 3 minus 2. Now, as a goes to infinity, this whole expression becomes infinity.

And so infinity minus 2 times 1 is just infinity. So if the integral test gives you infinity, that means that the original series diverges, which was in harmony with the other conclusion that we got based on the direct comparison test. Let's say if we have the series of the square root of n divided by n cubed plus 2. Which test should we use for this one?

The test that I recommend using is the limit comparison test. But we need two separate functions, one of which is... more simplified than the other. So for the other series, I'm just going to get rid of the two. So it becomes the square root of n divided by n cubed.

And according to the limit comparison test, if we take the limit as n goes to infinity of the ratio of the two sequences, if it's equal to a finite positive number, then both series will either converge or they will both diverge. Now we can reduce this to a p-series. So this is n to the 1 half divided by n cubed and so a half minus 3 is 2.5, negative 2.5.

And so this becomes 1 over n to the 2.5. So p is 2.5 and that's greater than 1. So according to the p-series test, this series converges. So if this passes the limit comparison test, it will also converge.

Let's find out. Now let's take the limit as n goes to infinity of a sub n divided by b sub n. So let's call this our a sub n.

And this expression is going to be our b sub n sequence. So this is going to be the limit as n goes to infinity, and then a sub n, that's the square root of n, divided by n to the third plus 2, and then divided by b sub n, which is the square root of n over n cubed. Perhaps you heard of the expression keep change flip. Keep the first fraction the same, change division to multiplication, flip the second fraction.

So we can rewrite this as times n to the third divided by the square root of n. So these two will cancel, and so we have the limit as n goes to infinity. of n cubed divided by n cubed plus 2. So let's multiply the top and the bottom by 1 over n cubed.

So if we do that, we're going to get the limit as n approaches infinity, and this is going to be 1 divided by 1 plus 2 times 1 over n cubed. So when n becomes very large, 1 over n cubed goes to 0. So it's 1 over 1 plus 2 times 0, which is 1 over 1, and that's 1. So this is equal to a finite positive number. Therefore, according to the limit comparison test, because this series converges, this one must converge as well. Now, consider the next example.

Let's say it's 3n squared minus 9 divided by 7n squared plus 4 raised to the n. Which tests should we use? If you see something raised to the n power, it's a good indication that we need to use the root test.

So recall that the root test is this. Take the limit as n goes to infinity of the nth root of the absolute value of the sequence a sub n. Now, if it's less than 1, the series converges. If it's greater than 1 or if it's infinity, the series diverges.

If it's equal to 1, then it's inconclusive. So if we take the limit as n goes into infinity of the nth root, which is the same as raising everything to the 1 over n power. n times 1 over n, those will cancel.

And so we have the limit as n goes to infinity of 3n squared minus 9 divided by 7n squared plus 4. Notice that the degree of the numerator is the same as that of the denominator. So you can see that the limit is going to be 3 over 7. But if you want to show your work, you can multiply the top and the bottom by 1 over n squared, or use L'Hopital's rule. Let's use L'Hopital's rule for this example.

So the derivative of 3n squared is 6n, and the derivative of 7n squared is 14n. So using L'Hopital's rule again. This is going to become 6 over 14, which reduces to 3 over 7. So this is the answer.

Now, 3 over 7 is less than 1. So based on the root test, the series converges. Now, let's work on our final example. And so we're going to have this series of 2 raised to the n divided by n factorial. So what tests should we use for this example?

So since you know it's the last problem, it's going to be the test that we haven't used yet, which is the ratio test. And so the basic idea behind a ratio test is we need to take... the a sub n plus 1, and divided by a sub n, all within an absolute value symbol. And if this is less than 1, the series converges.

If it's greater than 1 or equals infinity, the series diverges. Now, if it's equal to 1, then it's inconclusive. Now, we know that a sub n is 2 to the n over n factorial.

But what's a sub n plus 1? All you need to do is replace n with n plus 1. So the limit as n approaches infinity is going to be a sub n and then divided by, I mean, I'll take that back. It's a sub n plus 1. We need to write that one first. And then divided by a sub n.

Using keep change flip, we could say it's multiplied by 1 over a sub n, or the reciprocal of a sub n. So this is a sub n plus 1 times 1 over a sub n. So you can represent it that way if you want to.

So how can we simplify the limit now? What do we need to do? 2 to the n plus 1, that's 2 to the n times 2 to the first power.

n plus 1 factorial is n plus 1 times n factorial. And so we can cancel 2 to the n, and we can cancel n factorial. So we're left with the limit as n goes to infinity of 2 divided by n plus 1. 2 over n plus 1 is going to go to 0 as n goes to infinity. So 0 is less than 1, which means that the series converges by the ratio test.

And that's the answer.

Transcript for:Understanding Convergence and Divergence in Series

Transcript for:
Understanding Convergence and Divergence in Series