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How to find the sum of the series $\sum_1^\infty{\frac{1}{n(n+1)(n+2)}}$?

I expanded it via partial fractions but it does not look like a telescoping series which I was expecting.

Am I missing something obvious or easy manipulation here?

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    $\begingroup$ $$\frac{1}{n(n+1)(n+2)} = \frac{1}{2n}-\frac{1}{2(n+1)}+\frac{1}{2(n+2)}-\frac{1}{2(n+1)}$$ $\endgroup$ Jun 15, 2017 at 6:05
  • $\begingroup$ ... telescoping series $\;\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left(\frac{1}{n (n+1)} - \frac{1}{(n+1)(n+2)}\right)$ $\endgroup$
    – dxiv
    Jun 15, 2017 at 6:05
  • $\begingroup$ See also: math.stackexchange.com/questions/560816/… $\endgroup$ Jun 15, 2017 at 6:07
  • $\begingroup$ take limit to finite representation of the sum, which is $\tfrac{1}{4}-\tfrac{1}{2(n+1)(n+2)}$ $\endgroup$
    – serg_1
    Jun 15, 2017 at 6:50

1 Answer 1

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HINT:

$$\dfrac2{n(n+1)(n+2)}=\dfrac{n+2-n}{n(n+1)(n+2)}=\dfrac1{n(n+1)}-\dfrac1{(n+1)(n+2)}=f(n)-f(n+1)$$ where $f(m)=\dfrac1{m(m+1)}$

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