How to determine the convergence of $\sum_{n\geq 2}{} \frac{(-1)^{n}}{(-1)^n+n}$? This is most likely very easy to show, but with the load of midterms I had, my brain just declines to work properly. How do I determine the convergence of $$\sum_{n\geq 2}{} \frac{(-1)^{n}}{(-1)^n+n}$$?

HINT: 
$$\begin{align}
\sum_{n=2}^{2N}\frac{(-1)^n}{(-1)^n+n}&=\sum_{n=1}^{N}\left(\frac{1}{2n+1}-\frac{1}{2n}\right)+\frac{1}{2N}\\\\
&=\frac1{2N}-\sum_{n=1}^N \frac{1}{2n(2n+1)}
\end{align}$$

SPOILER ALERT:  Scroll over the highlighted area to reveal the solution

First, since $\left|\sum_{n=1}^N\frac{1}{2n(2n+1)}\right|\le \frac{1}{4}\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{24}$, we see that the series of interest converges.  In fact, we can evaluate the series in closed form.   Proceeding, we write $$\begin{align}\sum_{n=2}^{2N}\frac{(-1)^n}{(-1)^n+n}&=\sum_{n=1}^N\left(\frac{1}{2n+1}-\frac{1}{2n}\right)+\frac1{2N}\\\\&=\sum_{n=1}^N\left(\frac{1}{2n+1}+\frac{1}{2n}\right)-\sum_{n=1}^N\frac1n+\frac1{2N}\\\\&=-1+\sum_{n=1}^{2N+1}\frac1n-\sum_{n=1}^N\frac1n+\frac1{2N}\\\\&=-1+\sum_{n=1}^{N+1}\frac{1}{n+N}+\frac1{2N}\\\\&=-1+\frac1N\sum_{n=1}^{N+1}\frac{1}{1+n/N}+\frac{1}{2N} \tag {A1}\\\\&\to -1+\int_0^1 \frac{1}{1+x}\,dx\,\,\text{as}\,\,N\to \infty \tag{A2}\\\\&=-1+\log(2)\end{align}$$where we used only elementary arithmetic to take us to $(A1)$ and recognized the sum in $(A1)$ as a Riemann sum to arrive at $(A2)$.  An alternative way forward to evaluating the series is to write $$\sum_{n=1}^N\left(\frac{1}{2n+1}-\frac{1}{2n}\right)+\frac1{2N}=-1+\sum_{n=1}^{2N+1}\frac{(-1)^{n-1}}{n}+\frac{1}{2N}$$Then, recalling that $\log(1+x)$ has Taylor series representation $\log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}$ for $-1<x\le 1$, we see that $$\sum_{n=2}^\infty\frac{(-1)^n}{(-1)^n+n}=-1+\log(2)$$as expected!

