Is $\sum\limits_{n=0}^{\infty} \frac{n}{n+1} (-1)^{n}$ convergent? I will like to know whether $\sum\limits_{n=0}^{\infty} \frac{n}{n+1} x^{n}, \: x \in \mathbb{R}$ is convergent for $x=\pm1$ or not.
I found the radius of convergence to be $r=1$. So I have to check each ($x=1$ and $x=-1$) case separately. $x=1$ is easy. $x=-1$ is not so easy. I have used the limit test, M-test, root test, Leibniz test, comparison tes. But none if them work. I do not think I have other tests in my arsenal.

Any help is welcome.
You've already noted @Tom explained that, if $\sum_{n\ge0}a_n$ converges, the triangle inequality implies $\lim_{n\to\infty}a_n=0$. Whatever name that test has, it proves your series diverges. Indeed$$a_n:=\frac{(-1)^nn}{n+1}\implies\lim_{n\to\infty}|a_n|=1\implies\lim_{n\to\infty}a_n\ne0.$$In fact $\lim_{n\to\infty}a_n$ doesn't exist, because the subsequences for even (odd) $n$ have respective limits $1$ ($-1$), which differ. (That test appears to have a name too.)
OK, note that

$$\frac{-n}{n+1} + \frac{n+1}{n+2} = \frac{1}{(n+1)(n+2)}.$$   

So if you pair up the terms, you get a series with terms less than $1/n^2$.  So the series converges when $x=-1$.

Now if you pair them up the other way, you'll get a series that also converges to a different number.  So now what?