How can I show whether the series $\sum\limits_{n=1}^\infty \frac{(-1)^n}{n(2+(-1)^n)} $ converges or diverges? 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!
 A: Using the following proposition will help you resolve the problem:
If $a_n \longrightarrow_{n \to \infty} 0$ then $\sum_{n \ge 1} (-1)^n a_n$ and $\sum_{n \ge 1} (a_{2n}-a_{2n-1})$ either both converge or both diverge.
A: Hint:
$$\sum_{n=2k-1}^{2k} \frac{(-1)^n}{n(2+(-1)^n)}=-\frac{1}{2k-1}+\frac{1}{6k}=-\frac{4k+1}{6k(2k-1)}.$$
A: HINT: The Comparison Test is the gold standard for series that don't behave (like alternating series) and this leads to:
$$ \frac {1}{1+2^n} \ge \frac {1}{1+2^nn} \ge \frac {1}{1+(-2)^nn}\ge \frac {1}{1-2^nn} \ge \frac {1}{1-2^n}$$
Indeed,  the centre series (your series rearranged) alternates between the two series either side of it, making this a very safef comparison test as the bounds are so tight.
FULL ANSWER:
Using integral tests on the leftmost and rightmost fractions worked for me, although the integration was a bit of a pain:
$$ \sum_{n=1}^{\infty} \frac {1}{1+2^n} < \int_1^\infty \frac {1}{1+2^x}dx +1 =\frac {\ln(3/2)}{\ln 2}+1 \\\sum_{n=1}^{\infty} \frac {1}{1-2^n} < \int_1^\infty \frac {1}{1-2^x}dx -1 =-2 $$
Therefore both series converge and  thus your series converges (to -0.921454  with 6 decimal places.) 
