What is the sum of the series $$f(x)=\sum_{n=1}^\infty \frac {x^{n+2}}{n(n+2)}$$ in terms of $x$, where $-1 \le x \le 1$?
I can clean it up by differentiating:$$f'(x)=\sum_{n=1}^\infty \frac {x^{n+1}}n$$but I'm not sure where to go from here.
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lol, my anterior comment was terrible wrong, sorry :p. If you differentiate again you get something easy to compute. Because $f''$ is uniformly convergent for $x\in(-1,1)$ then you can compute from here $f$. Pull out an $x$ in $f'$ and differentiate the series. – Masacroso Mar 12 '17 at 08:28
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$$f'(x)=\sum_{n=1}^\infty \frac {x^{n+1}}n=x\sum_{n=1}^\infty \frac {x^n}n:=xg(x)\\ g'(x)=\sum_{n=1}^\infty {x^{n-1}}=\sum_{n=0}^\infty {x^n}=\frac 1{1-x},\;|x|<1$$ Can you continue from there?
jeckerya
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Hint. You can proceed with the following standard Taylor series expansion: $$ \sum_{n=1}^\infty \frac {x^{n}}n=-\ln(1-x),\quad |x|<1. $$
Olivier Oloa
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HINT: Try rewriting the series as: $$f(x)= \sum _{n=1}^{\infty}x^{n+2}\cdot\frac{1}{n(n+2)} = \frac{1}{2}\sum_{n=1}^{\infty}x^{n+2}\left(\frac{1}{n} -\frac{1}{n+2}\right)$$
Rohinb97
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