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How can I show whether the series $\sum\limits_{n=1}^\infty \frac{(-1)^n}{n(2+(-1)^n)} $ converges or diverges?

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While revising for exams, I came across a question where at the end of it, we had to determine whether the below series was convergent or divergent:

$$\sum_{n=1}^\infty \frac{(-1)^n}{n(2+(-1)^n)} $$

Unfortunately, other than knowing that as the series $ \sum_{n=1}^\infty \frac{(-1)^n}{n} $ converges, this series will most likely converge as the $(2 + (-1)^n)$ terms are bounded, I can't see a way of using this to determine if it converges or not.

Here's a list of some of the other ideas I've tried, which don't seem to get me anywhere:

  • Generalising the series - By this, I mean replacing the $-1$ for $z$ and then try and use the ratio test to see whether $-1$ is inside or outside or on the boundary of the circle of convergence - however, when I did so, I believe that the ratio test is inconclusive for all values of $|z|$, which while disappointing, is an interesting feature of the series.
  • Summation by parts - While this is usually a nice tool to use for when dealing with awkward sums, I can't see any good choice of sequences $ (a_n) $ and $(b_n)$ to choose to use.
  • Summing consecutive terms - By this, I mean evaluating the series by looking at the sum of

$$\sum_{k=1}^\infty \frac{(-1)^{2k}}{2k(2+(-1)^{2k})} + \frac{(-1)^{2k-1}}{(2k-1)(2+(-1)^{2k-1})}$$

$$ \Rightarrow \sum_{k=1}^\infty \frac{1}{2k} - \frac{1}{3(2k-1)}$$

$\quad$ but then as we are left with parts of the harmonic series this doesn't seem like a nice way to $\quad$evaluate it.

Other than that, I'm completely out of ideas, so any new ones (or ways to make my old ones work) would be much appreciated!


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