Find the set, where the series converges: $\sum_{n=1}^\infty\frac{(-1)^nn}{n^3+1}(x-3)^n$ I have to find the set, where the series converges: $$\sum_{n=1}^\infty\frac{(-1)^n n}{n^3+1}(x-3)^n.$$
I have already found $r=1$ and the center is $3$. So the set is $(2,4)$. Now I have to check, whether the series
is converges for $x=2$ and $x=4$.
And I don't know how to check it for $x=2$.
I don't know how to check is this series convergent: $$\sum_{n=1}^\infty\frac{(-1)^n n}{n^3+1}(-1)^n.$$
Please help me. I have been stuck here for 2 hours.

For both $x=2,4$, we have that
$$|(2-3)^n|=|(-1)^n|=1\text{ and }|(4-3)^n|=|1^n|=1$$
Then
$$\sum_{n=1}^\infty \left|\frac{(-1)^nn}{n^3+n}(x-3)^n\right|=\sum_{n=1}^\infty \frac{n}{n^3+n}=\sum_{n=1}^\infty \frac{1}{n^2+1}<\sum_{n=1}^\infty \frac{1}{n^2+0}=\sum_{n=1}^\infty \frac{1}{n^2}<\infty$$
(this last sum converges by the p-test).Thus, the original sum converges absolutely at $x=2,4$.
