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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.

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  • $\begingroup$ 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). $\endgroup$
    – hardmath
    Aug 16, 2018 at 12:51
  • $\begingroup$ 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 $n=2$? $\endgroup$
    – hardmath
    Aug 16, 2018 at 12:57
  • $\begingroup$ This is not a power series, so "interval of convergence" should be replaced by "domain of convergence" $\endgroup$
    – zhw.
    Aug 19, 2018 at 15:05

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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[$

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  • $\begingroup$ The series also converges at $y=-1.$ $\endgroup$
    – zhw.
    Aug 19, 2018 at 15:03
  • $\begingroup$ thanks, i edited my solution $\endgroup$
    – JayTuma
    Aug 19, 2018 at 15:50

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