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Show whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{(2+(-1)^n)n}$ converges (absolutely, conditionally) or diverges.

Edit: It's clear that the series can't be absolutely convergent since the series is greater than a constant multiple of the harmonic series which diverges. So I just have to check if the original series converges/diverges.

Arctic Char
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  • Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz Oct 08 '20 at 22:06

2 Answers2

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hint

$$|u_n|=\frac{1}{(2+(-1)^n)n}\ge \frac{1}{3n}$$

So, it is not absolutely convergent.

Think about $ u_{2n+1}+u_{2n}$.

hamam_Abdallah
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  • So I got $u_{2n+1} + u_{2n} = \frac{8n+1}{12n^2 + 6n}$. I'm trying to figure out how I can apply some series tests, but this is a sum of two $u$'s. –  Oct 08 '20 at 22:35
  • @AdamSmithSings I meant the separation between even and odd indices, as pointed below by User. – hamam_Abdallah Oct 08 '20 at 23:09
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We have that

$$\left| \frac{(-1)^n}{(2+(-1)^n)n}\right| \ge \frac{1}{3n}$$

and

$$\sum_{n=1}^{2N+1} \frac{(-1)^n}{(2+(-1)^n)n}=\sum_{n=1}^{N} \frac{1}{6n}-\sum_{n=0}^{N} \frac{1}{2n+1}=-1-\sum_{n=1}^{N} \frac{4n-1}{6n(2n+1)}$$

user
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  • What test would you use to show that $\sum_{n=1}^\infty \frac{4n-1}{6n(2n+1)}$ diverges? It should diverge because its similar to the harmonic series. –  Oct 08 '20 at 23:37
  • Yes as an intuition. Rigorously we can use direct comparison test or limit comparison test. – user Oct 09 '20 at 05:39