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Understanding Series Tests in Mathematics

Feb 25, 2025

Series Tests in Mathematics

Introduction to Series Tests

  • Today's focus: Direct Comparison Test (DCT) and Limit Comparison Test (LCT).
  • Both tests involve terms that are positive (greater than zero).

Direct Comparison Test (DCT)

  • Definition: If (a_n \leq b_n) for all or most terms:
    • If the series (\sum b_n) converges, then (\sum a_n) converges.
    • If (\sum a_n) diverges, then (\sum b_n) diverges.
  • Concept: Compare (a_n) to a known series (b_n).
    • If (b_n) converges and is larger, (a_n) must converge.
    • Consider that convergence is not heavily influenced by initial terms; focus on eventual terms.

Limit Comparison Test (LCT)

  • Definition: Focuses on the limit of the ratio of two series’ terms.
    • If (\lim_{n \to \infty} \frac{a_n}{b_n} = L) where (L) is finite and positive, both series either converge or diverge.
    • Special Cases:
      • If the limit is zero, both series converge.
      • If the limit is infinity, both series diverge.
  • Execution: Ensure (a_n) is compared correctly on top of (b_n)._

Applying the Tests

  • Selection of Comparison Series:
    • Use known series like geometric series or p-series for comparison.
    • Simplify terms to identify resemblance to known series.

Example Problems

Example 1: Use of DCT

  • Problem: Determine convergence of a series resembling (\frac{1}{n^3}).
  • Solution:
    • Known Series: (\frac{1}{n^3}) is a p-series with (p = 3) which converges.
    • Test: Show inequality by comparing the given series to (\frac{1}{n^3}).
    • Perform cross multiplication to establish (n^3(n-1) \leq n^4 + 2).
    • Conclusion: The given series converges by DCT.

Example 2: Use of LCT

  • Problem: Determine divergence of a series resembling (\frac{1}{\sqrt{n}}).
  • Initial Attempt (DCT):
    • Comparison Series: (\frac{1}{\sqrt{n}}) diverges.
    • Failed inequality: (\sqrt{n} \not\geq 2 + \sqrt{n}).
  • Switch to LCT:
    • Limit calculation: (\lim_{n \to \infty} \frac{\text{given}}{\frac{1}{\sqrt{n}}} = 1).
    • Conclusion: The given series diverges by LCT._

Conclusion

  • Both DCT and LCT are valuable in determining series convergence/divergence.
  • Essential to practice choosing appropriate comparison series.
  • Remember: It’s often necessary to switch tests if one fails.
  • Practice using known series and working through inequalities or limits to verify results.