Determine if $\sum_{n=2}^{\infty} \frac{(-1)^n}{n+(-1)^n}$ converges or diverges I'm having a lot of trouble figuring this one out.
Determine if  $\sum_{i=2}^{\infty} \frac{(-1)^n}{n+(-1)^n}$ converges or diverges
Both ratio and root test are inconclusive and I'm at a loss. Can anyone help me? 
 A: Let
$$S_n=\sum_{k=2}^n \frac{(-1)^n}{n+(-1)^n}$$
Take the sum of two consecutive terms:
$$\frac{(-1)^{2n}}{2n+(-1)^{2n}}+\frac{(-1)^{2n+1}}{2n+1+(-1)^{2n+1}}=\frac{1}{2n+1}-\frac{1}{2n}=\frac{-1}{2n(2n+1)}$$
Thus $S_{2n+1}$ converges. Since the terms tend to $0$, $S_{2n}$ is also converging, to the same limit.
A: One may observe that, as $n \to \infty$, we have
$$
\begin{align}
\frac{(-1)^n}{n+(-1)^n}&=\frac{(-1)^n}{n}\times\frac1{1+\frac{(-1)^n}{n}}\\\\
&=\frac{(-1)^n}{n}\left(1-\frac{(-1)^n}{n}+\mathcal{O}\left(\frac1{n^2} \right)\right)\\\\
&=\frac{(-1)^n}{n}-\frac1{n^2}+\mathcal{O}\left(\frac1{n^3} \right)
\end{align}
$$ then, for some integer $p$,

$$
\underbrace{\sum_{n\geq p}\frac{(-1)^n}{n+(-1)^n}}_{{\color{red}{\text{conditionally CV}}}}=\underbrace{\sum_{n\geq p}\frac{(-1)^n}{n}}_{{\color{red}{\text{conditionally CV}}}}-\underbrace{\sum_{n\geq p}\frac1{n^2}+\sum_{n\geq p}\mathcal{O}\left(\frac1{n^3} \right)}_{{\color{blue}{\text{absolutely CV}}}}
$$ 

and your initial series is conditionally convergent.
A: Let's write
$$\frac{(-1)^n}{n+(-1)^n} = \left ( \frac{(-1)^n}{n+(-1)^n} - \frac{(-1)^n} {n} \right) + \frac{(-1)^n}{n}.$$
The term in parentheses equals $-1/[(n+(-1)^n)n].$ In absolute value, these terms are $\le 1/(n-1)^2.$ Hence the series of these terms converges absolutely. The series $\sum (-1)^n/n$ converges by the alternating  series test. So our series is the sum of two convergent series, hence converges.
A: Terms are alternatively positive and negative, decrease in absolute value and converge to $0$, so the sseries converges (this is the alternating series test), but it does not converge absolutely.
