I am having difficulty finding if this series converges or diverges: ∑n=0∞(−2)n+1nn−1 I am unsure of which test to use. At first, I thought I should use alternating series test, but I am unable to manipulate the numerator to (−1)n+1. I thought I may be able to use root test, but I have no clue how to manipulate the series to do that. What series test would I use, and how?

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I am having difficulty finding if this series converges or diverges:  ∑n=0∞(−2)n+1nn−1  I am unsure of which test to use. At first, I thought I should use alternating series test, but I am unable to manipulate the numerator to (−1)n+1. I thought I may be able to use root test, but I have no clue how to manipulate the series to do that.  What series test would I use, and how?

Ethan Sanders · Accepted Answer

Alternating series test  Note that  ∑n=0∞(−2)n+1nn−1=∑n=0∞(−1)n+12n+1nn−1  By the alternating series test this converges if  1. an&amp;gt;0: we can easily this is satisfied  2. an is monotic: an is decreasing monotonically  3. limn→∞an=0 I will leave this one to you.  Comparison test  |∑n=0∞(−2)n+1nn−1|≤∑n=0∞2n+1nn−1≤∑n=03(2n)n+1+∑n=4∞(2n)n+1  The last series is a geometric series and therefore converges.

Virginia Palmer · Accepted Answer

Hint: what can you say about 2n−1nn−1?  Can you show that 2n−1nn−1≤1n2 for n≥N?

user_27qwe · Accepted Answer

Making the problem more general, let S=∑n=0∞an with an=xn+1nn−1 log⁡(an)=(n+1)log⁡(x)−(n−1)log⁡(n) log⁡(an+1)−log⁡(an)=(n−1)log⁡(n)−nlog⁡(n+1)+log⁡(x) Using Taylor for large values of n log⁡(an+1)−log⁡(an)=log⁡(xne)+12n+O(1n2) an+1an=elog⁡(an+1)−log⁡(an)∼xne→0

I am having difficulty finding if this series converges or d

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