# What is the interval of convergence for power series

I have a power series and I need to find the interval of convergence. $$\sum_{n=0}^\infty= \frac {n}{(n^2-1) (1+x)^n}$$ I tried ratio test with a new variable $t = \frac{1}{1+x}$and I get that radius of convergence is one ( $R= 1$) so the interval of convergence is $x \in [-2,0]$. But it doesn't seem right, because if I put $x=4$ by ratio test this converges. So my question is what is the interval of convergence? And how do you get it? Thank you.

• You got convergence for $t\in (-1,+1)$, but we need to investigate the endpoints for convergence also. In any case the change of variable from $t$ back to $x$ is not simply a translation (because $t$ involves the reciprocal). – hardmath Aug 16 '18 at 12:51
• Note as a technical detail that the series summation includes a term $n=1$ which involves division by zero in a coefficient. Perhaps you should start the summation at index $n=2$? – hardmath Aug 16 '18 at 12:57
• This is not a power series, so "interval of convergence" should be replaced by "domain of convergence" – zhw. Aug 19 '18 at 15:05

Just consider another power seriese $$g(y) = \sum_{n = 2}^\infty \frac{n}{n^2 - 1} \cdot y^n$$ This is a standard power series and converges in $[-1,1[$. Now to determine where your series converges, just notice that you need to apply the change of variable $\displaystyle y = \frac{1}{x+1}$ so the condition $y \in [-1,1[$ is equivalent to $x \in \; ]-\infty, -2] \; \cup \; ]0, +\infty[$
• The series also converges at $y=-1.$ – zhw. Aug 19 '18 at 15:03