Convergence of alternating series (not monotone) I'm dealing with the following series
$$\sum_{n\ge 2} \frac{1}{2+(-1)^nn}\,.$$
Clearly this will not converge absolutely as the general term in absolute value $\sim 1/n$. Leibniz can not be applied as we do not have the monotonicity by a hair.
This one has to be converging as it is behaving almost as $(-1)^n/n$. Point is, how do I show it? Any idea? 

Following the idea of Qiaochu, whom I thank, we have
$$\frac{1}{2+(-1)^nn} - \frac{(-1)^n}{n} = \frac{n-2(-1)^n-n}{2n+(-1)^nn^2} = -\frac{2}{2n+(-1)^nn^2}$$
Then
$$\frac{1}{2+(-1)^nn} = \frac{(-1)^n}{n}-\frac{2}{2n+(-1)^nn^2}\,.$$
As the series corresponding to the terms of the RHS converge as the first one is $(=1)^n/n$ and the second one goes as $(-1)^n/n^2$, so does the one corresponding to the LHS.

Possibly another way to do it. Get rid for free of the first term, then consider the series from $n\ge 3$.
$$\frac{1}{2+(-1)^nn} = \frac{1}{2-(-1)^{n+1}n} = \frac{1}{2-(-1)^{n+1}n} \cdot \frac{2+(-1)^{n+1}n}{2+(-1)^{n+1}n} = \frac{2+(-1)^{n+1}n}{4-n^2}\,.$$
Hence we can break it as
$$\frac{2}{4-n^2} + \frac{(-1)^{n+1}n}{4-n^2}\,.$$
The first piece converges absolutely, while the second conditionally.
