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Alternating Series Calculator

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Example
Created on 2024-06-20Asked by Camila Adams (Solvelet student)
Determine if the alternating series ∑n=1∞(−1)nn2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} converges.

Solution

Step 1: Identify the series and the general term. The series is ∑n=1∞(−1)nn2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} with general term an=(−1)nn2. a_n = \frac{(-1)^n}{n^2}. Step 2: Apply the Alternating Series Test (Leibniz's test), which requires: \begin{enumerate} \item bn=1n2b_n = \frac{1}{n^2} is positive, \item bnb_n is decreasing, \item lim⁡n→∞bn=0\lim_{n \to \infty} b_n = 0. \end{enumerate} Step 3: Verify the conditions: \begin{enumerate} \item bn=1n2b_n = \frac{1}{n^2} is positive for all n≥1n \geq 1. \item bn=1n2b_n = \frac{1}{n^2} is decreasing because for n≥1n \geq 1, 1n2>1(n+1)2\frac{1}{n^2} > \frac{1}{(n+1)^2}. \item lim⁡n→∞1n2=0\lim_{n \to \infty} \frac{1}{n^2} = 0. \end{enumerate} Step 4: Conclusion. Since bnb_n is positive, decreasing, and approaches zero as n→∞n \to \infty, the series ∑n=1∞(−1)nn2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} converges by the Alternating Series Test. Solved on Solvelet with Basic AI Model
Some of the related questions asked by Henry Smith on Solvelet
1. Determine the convergence of the series ∑n=1∞(−1)n/n2 \sum_{n=1}^{\infty} (-1)^n / n^2 2. Determine the convergence of the series ∑n=1∞(−1)n2n+1 \sum_{n=1}^{\infty} \frac{(-1)^n}{2n + 1} .,
DefinitionAn alternating series is one for which the terms are alternately positive and negative; this is often written as ∑(−1)nan​ or ∑(−1)n+1an​. Series of such kind may converge in some cases. EX/THE Alternating Series Test (Leibniz's Test) If an​ decrease monotonically to 0, then the series ∑(−1)nan​ converges. Practice solving Alternating Series questions with SolveletAI advanced step by step solutions. Instantly generated, Having a hard time with the problem explanation - Alternating Series calculator at SolveletAI.
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