Proving whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. I've updated my proof to be complete now, edited for proof-verification!

We know that for the partial sums with even an uneven terms, the following holds:
$S_{2N}=\sum_{n=1}^{2N} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3} -\frac{1}{6} +\frac{1}{5}-\dots+\frac{1}{2N-1}$
$= \frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$
We may rewrite the series in pairs as we know  it will have an even amount of terms.
$S_{2N+1} = \sum_{n=1}^{2N+1} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3}-\dots+\frac{1}{2N-1} -\frac{1}{2N+2}$
$=\frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} - \frac{1}{2N+2} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}-\frac{1}{2N+2}$
As $n\in\mathbb{N}$, we know that $n\geq1$ so:
$n\geq1 \iff 3n\geq3 \iff 3n^2\geq3n \iff 3n^2-3n\geq0 \iff 4n^2-2n \geq n^2+n$
So: $2n(2n-1)\geq n(n+1) \iff \frac{1}{2n(2n-1)}\leq \frac{1}{n(n+1)}$ for all $n\geq1$. 
As the series of the latter sequence converges, we can conclude, by the comparison, test that the series $\sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$ converges. 
Suppose it converges to $s$, then we know $^{\lim S_{2N}}_{N\to\infty} = s$ and thus $\lim_{N\to\infty}[S_{2N+1}] = s - (\lim_{N\to\infty}[\frac{1}{2N+2}]) = s-0 = s.$ As the partial sums ending with even and uneven terms both converge to the same limit, the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$  converges. $\tag*{$\Box$}$ 

 A: \begin{align*}
\sum_{k=2}^{N}\dfrac{1}{2k(2k-1)}\leq\sum_{k=2}^{N}\dfrac{1}{2k(2k-(k/2))}=\dfrac{1}{3}\sum_{k=2}^{N}\dfrac{1}{k^{2}}<\dfrac{1}{3}\sum_{k=2}^{\infty}\dfrac{1}{k^{2}}<\infty,
\end{align*}
so $\{S_{2N}\}$ is convergent, so is $\{S_{2N+1}\}$ because $\lim_{N}(S_{2N+1}-S_{2N})=0$ (so they have the same limit). Then $\{S_{N}\}$ is convergent.
A: As you computed, it's the alternating harmonic series in a different order. In fact the first N terms are just the first N terms of the alternating harmonic series permuted (edit: with an extra term which decays like 1/N in case N is odd), so the sequence of partial sums is the same.
A: Another way is to observe that
$$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n-(-1)^{n}}=\sum_{n=1}^{\infty}\Bigg[\frac{(-1)^{n}}{n-(-1)^{n}}-\frac{(-1)^{n}}{n}\Bigg]+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}=$$
$$=\sum_{n=1}^{\infty}\frac{1}{n^{2}-(-1)^{n}n}+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$$
Now the left series converges by the limit comparison test with $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$, and the right series converges by Leibniz criterion.
A: Let:
$$a_n=\frac{1}{n-(-1)^n}$$
Note that:


*

*for $n=2k-1 \implies a_{2k-1}=\frac{1}{2k}$

*for $n=2k \implies a_{2k}=\frac{1}{2k-1}$

*for $n=2k+1 \implies a_{2k+1}=\frac{1}{2k+2}$

*for $n=2k+2 \implies a_{2k+2}=\frac{1}{2k+1}$


Therefore we can reorder the series $a_n\to b_n$ in such way that $b_n$ is monolitically decreasing, since:
$$\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}=\sum_{n=1}^\infty (-1)^na_n=\sum_{n=1}^\infty (-1)^nb_n<+\infty$$
the series converges.
A: $$s =\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n} = $$
$$ -\frac{1}{2} + \frac{1}{1} - \frac{1}{4}  + \frac{1}{3} - \frac{1}{6} + \frac{1}{5} - \cdots $$
$$\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\cdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)$$
$$ \sum_{k =1}^\infty \left(\frac{1}{2k-1}\right) - \sum_{k =1}^\infty \left(\frac{1}{2k}\right)$$
$$ \sum_{k =1}^\infty \left(\frac{1}{2k-1} - \frac{1}{2k}\right)$$
$$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}$$

$$s = \ln(2)$$

And the series converges to the natural logarithm. 

See the special case of the Mercator series 
 $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
Setting $x=1$ in the Mercator series yields the Alternating harmonic series
$$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}=\ln(2)$$
