okay hello everyone welcome back to another video today we're going to be tackling a particularly tricky je problem and for anyone who doesn't know it's rowned to be one of the hardest uh maths exams Sat by students around the world uh it's taken in India and I've heard from lots of sources online it's kind of regarded to be one of the hardest questions that's ever come up in recent years so let's see if we can give it our best shot it gives us a sum SN and defines it as the sum from K = 1 to n of n / n^ 2 + k n + k^ 2 and this is for n in the natural numbers 1 2 3 Etc and it gives us two inequalities here to look at it also gives us two other inequalities which I might leave as a challenge at the end but we're going to stick to this main part of the question for today and it tells us to determine whether either s of n is less than pi over 3 < tk3 or whether it's greater than pi over 3 < tk3 and I love this question because it's so difficult to know where to start we start seeing pies appear and it just looks like a disaster where do we even go to get a pi and an inequality like that but what we're going to first consider is how could we bound SN what what would be the maximum value this sum could take and could we create an inequality there well let's think about what we're summing right it's this it's from 1 to n of n / N2 + k n + k 2 and this this value you this series will never have a negative term in it because there are no negative numbers and there's nothing negative involved in our sum which means that every time we increase n every time we add one more term even if it's adding a tiny tiny bit it will be making SN slightly bigger and so that means that the largest thing that SN could possibly be would be when we let N approach Infinity it would be the limit as n approached Infinity of SN and so we can confidently say that SN will always be less than S infinity and so if we could work out what s Infinity is and do something with that perhaps that's going to lead us to the answer we're looking for so let's consider that limit as n goes to Infinity of the sum from K = 1 to n or n / n^2 + k n + k^ 2 well the first thing I'm going to do is factor out an N everywhere that I'm looking cuz I always get a bit uncomfortable when we have things blowing up to Infinity on our numerator and that's going to lead us to the Limit as n goes to Infinity of the sum from K = 1 to n of one / by well we can cancel an N there we can cancel an N there and we can't cancel an N there so we'll just divide by n like that okay well that's promising but but we're still adding an Infinity here so I'm not quite happy what I'm going to do is I'm going to divide everything by n again and see where that gets us so this is equal to the Limit as n approaches Infinity of the sum from K = 1 to n of 1 / n * by 1 / 1 + k / n + Well k^ 2/ n^ 2 also known as k/ n 2 and this is where the magic happens because there's something we've got to spot in this expression and this is what makes it one of the hardest je problems that's ever come up so I'm going to leave this here and actually I'm going to take a side step in our video and I'm going to remind everyone of reman sums what is a rean sum well it tells us that the integral from A to B of A given function of X with respect to X is the same same thing as taking the limit as n goes to Infinity of the sum from K = 1 to n of f of x k Times by Delta X now what's Delta X well Delta X is what we want our infant tmal change to be and there's a geometric inter interpretation for all of this it's that we're taking the limit from two points A and B of A given function and what we're choosing to do is we're defining some bars of equal width and the width of These Bars is Delta X and what we want to do is sum up these bars as they decrease in width to infin infinite decimal bars so we're going to we want that change in X along the x- axis to be that distance B minus a between the two bounds divided by n which is approaching Infinity so that will split evenly that distance B minus a into infinitees strips and of course if we want to find the area we're going to have to multiply the width on the x-axis by the height on the Y AIS and if we return to our graph again that means we're going to start at a and let's say we wanted to find the height for a bar four away from our starting point well what we want to do is add four multiples of the width of the bar to get along to our point and then input that into our function to find our height at that given point and so that's why we input into our function x k which is equal to our initial starting point a plus some multiple K of our change in X it moves us along exactly to where we need to be to find the height at given point and then we insert it into our function and the reason that this is important is because what we're looking at here is a formula for a rean sum we're looking at what 1 / n and that must be our Delta X which means that b - A = 1 and we've got our function here which looks to be of the form 1 over 1 + x + X2 and what we're inputting is x = k / n which must be our value here a + k Delta X so therefore our Delta X as we've already said must be 1 / n and our a must be Z which means that our B must be one and the reason that we know that is because if x k is equal to k / n and we're not adding any constant starting term we must have started at the origin so a is zero and if B minus a is equal to 1 then that means b - 0 is 1 which means B must be one so what we're going to say is that we know our bounds our bounds are starting from zero and going to one and summing rectangles of of of increasingly small width and therefore increasingly narrow area for a given function and that function we've managed to deduce is one over 1 + x + x^2 and so this is what makes such a tricky problem you have to be really comfortable with your rean sums and so we can now say confidently that SN is less than s infinity and at Infinity that sum is equivalent to the integral from 0 to 1 of 1 /x^2 + x + 1 with respect to X now many of you who have come across lots of integral problems before might recognize this instantly as an arctan integral what we want to do here is complete the square and by doing that we're going to end up with x + a half SAR + one minus a qu which is + 3 over 4 with respect to X and now all we have to do is use the arctan integral formula the arcan integral formula tells us that the integral of 1 / a^2 + U ^2 with respect to U is equal to 1 / a * R tan of u/ a plus c and of course I'll I can leave you guys to prove that for yourselves in the comments but it definitely works you can prove it by some nice substitutions and what's that going to leave us with this will being equal to well it's bounded so we don't have a plus c but it will be equal to well we've got to determine which part is the constant which part is our a and it's going to be < tk3 / 2^ 2 here which is our 3 over 4 and we're going to use 1 over a which is going to be 2 over < tk3 multiplied by R tan of our X component here divided by our constant component here which is going to be X Plus half / < tk3 / 2 and of course if I multiply by two both sides here I'm going to get 2x + 1 / < tk3 and let's not forget this is bounded between x = 1 and x = 0 now what does this leave us with well when X is equal to 1 we're going to be looking at 2/ < tk3 * R tan of 2 + 1 which is 3/ < tk3 in another words arctan of < tk3 and when X is equal to zero we're going to be taking Aran of 1 /3 now anyone preparing for the je or who's come across lots of trig before will recognize this is 2 over > 3 ultied > / 3 - > / 6 which = 2K3 * < / 6 which is < over 3 < tk3 and so we've Clearly Now shown that s of n is less than S of infinity and S of infinity is equal to < over 3 < tk3 yielding SN is less than < over 3 < tk3 I hope you've enjoyed this this question I thought it was a really nice one that involves some good rean sums and nice problem solving to come to a very satisfying answer at the end and I did mention a challenge at the end of the video and it defines a second sum in the subsequent part of the problem T of n which is equal to the sum of the same expression but starting from zero and ending with n minus one and it asks you to find or look at similar inequalities is T of n less than or greater than pi over 3 ro3 and I'll leave that as a challenge to you guys in the comments thanks a lot for watching and I'll see you next time bye