The region of convergence of the series $\sum_{n=1}^{\infty} \frac{(-1)^n n (x-1)^n}{(n^3+1)(3^n+1)}$.

Question

I am trying to find the region of convergence of the series $$ \sum_{n=1}^{\infty} \frac{(-1)^n n (x-1)^n}{(n^3+1)(3^n+1)}. $$ I know that, by the limit comparison test, $$ \sum_{n=1}^{\infty} \frac{n (x-1)^n}{(n^3+1)(3^n+1)} $$ converges if and only if $$ \sum_{n=1}^{\infty} \frac{(x-1)^n}{n^2 3^n} $$ converges, and, by the Leibniz criterion, $$ \sum_{n=1}^{\infty} \frac{(-1)^n (x-1)^n}{n^2 3^n} $$ converges if $|x-1|\le 3$. But I do not know how the convergence of the original series follows from these two facts.

Answer 1

https://en.wikipedia.org/wiki/Power_series check "Radius of convergence" here your $a_n=\frac{n(-1)^n}{\left(n^3+1\right)(3^n+1)}$. You just find $r=3$ but notice that your center here is $x_0=1$(and not $x_0=0$) so you demand $|x-1|<3$ and you check in the case $|x-1|=3$ if the series converges.