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Evaluate $$\sum_{i=2}^{\infty}{\frac{n}{n^2-1}}x^n$$ using the fact that

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

So far I have proven that the Radius of Convergence is 1 and that the series converges absolutely if $|x|<1$. I have looked at examples of evaluating telescoping sums, but that isn't applicable. Any help is greatly appreciated, as I'm kind of stumped on this part.

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  • $\begingroup$ Hint: $$ \sum\limits_{n = 2}^\infty {\frac{{x^n }}{{2(n - 1)}}} = \frac{x}{2}\sum\limits_{n = 2}^\infty {\frac{{x^{n - 1} }}{{n - 1}}} = \frac{x}{2}\sum\limits_{n = 1}^\infty {\frac{{x^n }}{n}} $$ and so on. Do you know the Maclaurin series of $-\log(1-x)$? $\endgroup$
    – Gary
    Commented Aug 19, 2021 at 17:31
  • $\begingroup$ Note $\sum_{n=1}^\infty \frac{1}{n} x^n = -\log(1-x)$. Then you can multiply/ divide your sum by $x$ to get the indices to match up so that the terms are in that form. $\endgroup$
    – Jair Taylor
    Commented Aug 19, 2021 at 17:33
  • $\begingroup$ I don't know that, but I will take a look at it now. Thank you all for the hints! $\endgroup$
    – mathnoob
    Commented Aug 19, 2021 at 17:34
  • $\begingroup$ when I changed the index from $\sum_{i=2}^{\infty}{\frac{1}{n+1}}x^n$ to ${\frac{x}{2}}\sum_{i=1}^{\infty}{\frac{1}{n+2}}x^n$ would this give me ${\frac{x}{2}}\log(2-x)$? $\endgroup$
    – mathnoob
    Commented Aug 19, 2021 at 18:56

1 Answer 1

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Hint : We can integrate and differentiate previously known decompositions of elementary functions

enter link description here

As a result, we get

$$ -\frac{x(2+x)+2(1+x^{2})\log (1-x)}{4x} $$

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