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

Solution 1:

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.

Solution 2:

Hint: $$\sum_{n=2k-1}^{2k} \frac{(-1)^n}{n(2+(-1)^n)}=-\frac{1}{2k-1}+\frac{1}{6k}=-\frac{4k+1}{6k(2k-1)}.$$

Solution 3:

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.)