∑n=1^∞ (-1)^n-1( (1 · 3 · 5 ⋯ (2n-1))/(2 · 4 · 6 ⋯ (2n)))^x

Name: ∑n=1^∞ (-1)^n-1( (1 · 3 · 5 ⋯ (2n-1))/(2 · 4 · 6 ⋯ (2n)))^x
Uploaded: 2022-05-30T22:47:13-08:00
Duration: 1 min 59 s
Channel: Zhumagali Shomanov
Description: ∑n=1^∞ (-1)^n-1( (1 · 3 · 5 ⋯ (2n-1))/(2 · 4 · 6 ⋯ (2n)))^x

Question

$\sum_{n=1}^{\infty} (-1)^{n-1} \left( \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \right)^x$

          $\sum_{n=1}^{\infty} (-1)^{n-1} \left( \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \right)^x$

∑n=1^∞ (-1)^n-1( (1 · 3 · 5 ⋯ (2n-1))/(2 · 4 · 6 ⋯ (2n)))^x

Added by Brian C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Zhumagali Shomanov

05/30/2022

Step 1

Step 1: Use the ratio test to find the values of α for which the series converges. Show more…

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sum_(n=1)^(infty ) (-1)^(n-1)((p*3*5dots(2n-1))/(2*4*6dots(2n)))^(alpha )Find all a values at which the series converges conditionally

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Question

$\sum_{n=1}^{\infty} (-1)^{n-1} \left( \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \right)^x$

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