Determine convergence of $\sum\limits_{k=1}^{\infty}\frac{(-1)^k k}{k^2-(-1)^k}$ I want to determine whether the following series converges:
$$\sum\limits_{k=1}^{\infty}\frac{(-1)^k k}{k^2-(-1)^k}$$
I managed to prove that it does not converge absolutely, however I fail to prove that the series converge conditionally (which seems to be the case). The Alternating series test doesn't work  (because $|a_1|<|a_2|$).
 A: $$\sum_{k=1}^{2N}\frac{(-1)^k k}{k^2-(-1)^k} = \sum_{h=1}^{N}\frac{2h}{4h^2-1}-\sum_{k=1}^{N}\frac{2h-1}{(2h-1)^2+1}\\=-\sum_{h=1}^{N}\frac{4h^2-6 h+1}{2 (2h-1) (2h+1) \left(2h^2-2 h+1\right)} $$
and the last one is a converging series, since the main term behaves like $\frac{1}{4h^2}$. By partial fraction decomposition, the value of the series is given by:
$$ -\frac{1}{2}+\sum_{n\geq 0}\frac{1}{2(2n+1)(2n^2+2n+1)}\approx 0.0477$$
that can be written as:
$$ \left(\frac{7}{8}\zeta(3)-1\right)-\sum_{n\geq 1}\frac{1}{(2n+1)^3(4n^2+4n+2)}.$$
A: Using alternating series test. 
1) prove $\lim\limits_{n\to\infty}a_n=0$
$$a_n=\frac n{n^2-(-1)^n}=\frac {1/n}{1-\frac{(-1)^n}{n^2}} \rightarrow 0$$
2) prove $a_n \gt a_{n+1}$
$$a_n=\frac n{n^2-(-1)^n}=\frac 1{n-\frac{(-1)^n}{n}}$$
$$a_{n+1}=\frac 1{(n+1)-\frac{(-1)^{n+1}}{n+1}}=\frac 1{n-\frac{(-1)^n}n+ (1 -\frac{(-1)^{n+1}}{n+1}+\frac{(-1)^n}n)}$$
when $n \gt 10$
$$1 -\frac{(-1)^{n+1}}{n+1}+\frac{(-1)^n}n>1-1/10-1/10 =4/5$$
$$a_{n+1}=\frac 1{n-\frac{(-1)^n}n+ (1 -\frac{(-1)^{n+1}}{n+1}+\frac{(-1)^n}n)}<\frac 1{n-\frac{(-1)^n}n+ 4/5}<\frac 1{n-\frac{(-1)^n}n}=a_n$$
DONE
