I need help finding the sum of the infinite series $$\sum_{k=1}^\infty \frac{1}{n(n+1)(n+2)}$$ I have used the partial fraction decomposition to get this as the sum of $$\frac{-1}{k+1}+\frac{1}{2(k+2)}+\frac{1}{2k}$$ but don't know where to go from here. Thanks!
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1$$\frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+1}\right) - \frac{1}{2}\left(\frac{1}{k+1} - \frac{1}{k+2}\right)$$ – Daniel Fischer Oct 24 '14 at 20:32
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But what if the sum doesn't telescope ? :-$)$ – Lucian Oct 24 '14 at 22:25
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@Lucian Partial fraction decomposition and digamma function always works :) – Olivier Oloa Oct 24 '14 at 22:52
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Hint: Easier is to use $$ \frac1{n(n+1)}-\frac1{(n+1)(n+2)}=\frac2{n(n+1)(n+2)} $$ and just add. Note that this is a telescoping series.
robjohn
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Hint. First observe that $$ \frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right) $$ and conclude by telescoping terms. You end up with $$ \sum_{n=1}^{N}\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(1-\frac{1}{N+1}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{N+2}\right) $$ and then let tend $N$ to $+\infty$.
Olivier Oloa
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