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?

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.

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.