<s>[INST] Show convergence by using partial fractions Show that $\sum\limits_{n=1}^{\infty}{\frac{1}{n(n+1)(n+2)}}=\frac{1}{4}$
Answer: 
$\sum\limits_{n=1}^{\infty}{\frac{1}{n(n+1)(n+2)}}=$ $\sum\limits_{n=1}^{\infty}{(\frac{1}{2n}-\frac{1}{n+1} + \frac{1}{2(n+2)})}$
= $1/2 -1/2 + 1/6 +1/4 -1/3 +1/8+1/6 -1/4 +1/10 +1/8 -1/5 +1/12 +....$
How do I proceed further? Can anyone give hints/guide?
Thank you.
 [/INST] We can rewrite the sum as
$$ \sum_{n=1}^{\infty} \frac{1}{2}\left( \left[ \frac{1}{n} - \frac{1}{n+1} \right] + \left[ \frac{1}{n+2} - \frac{1}{n + 1} \right] \right) $$
From this the way the sum telescopes should be clear.
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