Absolutely convergent, conditionally convergent or divergent series Show if the following series
$$
\sum_{n=2}^{\infty}\frac{(-1)^n}{(-1)^n + n}
$$
are absolutely convergent, conditionally convergent or divergent.
I think I succeeded in proving that it is NOT absolutely convergent. My attempt:
$$
\sum_{n=2}^{\infty}\left|\frac{(-1)^n}{(-1)^n + n}\right| = \sum_{n=2}^{\infty}\frac{(-1)^n}{\left|(-1)^n + n\right|}
$$
Then we have
$$
\lim_{n\to\infty}\frac{\frac{(-1)^n}{\left|(-1)^n + n\right|}}{\frac{1}{n}} = \lim_{n\to\infty}\frac{n}{(-1)^n+n} = 1
$$
Therefore
$$
\sum_{n=2}^{\infty}\frac{(-1)^n}{\left|(-1)^n + n\right|}\ \text{is convergent}\ <=> \sum_{n=2}^{\infty}\frac{1}{n}\ \text{is convergent.}\
$$
The last series are NOT convergent, so our series are NOT absolutely convergent.
I'm still stuck on proving if it is conditionally convergent or divergent. intuitively, I think It's divergent, but I need a formal proof. It crossed my mind that I can show $a_n = \frac{1}{(-1)^n +n}$ is NOT monotonously decreasing so It's divergent(Leibniz). But I think Leibniz's theorem doesn't work in this direction. So I'm still stuck.
Looking forward for any ideas!
 A: The series is not absolutely convergent, since$$\lim_{n\to\infty}\frac{\left|\frac{(-1)^n}{(-1)^n+n}\right|}{\frac1n}=\lim_{n\to\infty}\frac n{(-1)^n+n}=1$$and the harmonic series diverges.
On the other hand, your series is$$\frac13-\frac12+\frac15-\frac14+\frac17-\frac16+\cdots\tag1$$and you know, from the Leibniz test, that the series$$-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\cdots$$converges. It is not hard to deduce from this that the series $(1)$ converges too.
A: 
I haven't checked all the details but I think this works:
\begin{aligned} \sum_{n=2}^\infty
 \frac{(-1)^n}{(-1)^n+n}&=\frac13-\frac12+\frac15-\frac14+\cdots\\
&=\sum_{n=1}^\infty\frac{1}{2n+1}-\frac{1}{2n}\\
 &=\sum_{n=1}^\infty\frac{-1}{(2n+1)(2n)} \end{aligned} and you can use
the comparison test with $\frac{1}{n^2}$ to show that the last series
converges.

Thank you to @JoséCarlosSantos for pointing out that the above reasoning is incorrect. I leave it here in case someone else might come up with a similar "solution". A correct approach (I hope :)) is noting that
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=1-\frac12+\frac13-\frac14+\cdots=\ln2$$
and so
$$-1+\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=-\frac12+\frac13-\frac14+\cdots=\ln2-1.$$
And since the partial sums
$$S_N=-1+\sum_{n=1}^N\frac{(-1)^{n+1}}{n}=\sum_{n=2}^N\frac{(-1)^n}{(-1)^n+n}$$
agree, the original series converges (conditionally) to $\ln2-1$ as well.
