<s>[INST] Help with infinite sum Can you guys give me a hint on evaluating $$\sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)}?$$
I have tried partial fractions but the series is not telescopic (at least I cannot see it)...
 [/INST] Hint:

$$\frac{1}{n(n+2)(n+4)}=\frac{1}{4}\frac{n+4-n}{n(n+2)(n+4)}=\frac{1}{4}[\frac{1}{n(n+2)}-\frac{1}{(n+2)(n+4)}]$$

In general:
$$\frac{1}{n(n+d)...(n+kd)}=\frac{1}{kd}\left[\frac{1}{n(n+d)...(n+(k-1)d)}-\frac{1}{(n+d)(n+2d)...(n+kd)}\right]$$
Now if you let $a_n=\frac{-1}{kd}(\frac{1}{n(n+d)...(n+(k-1)d)})$, we find that:
$$\frac{1}{n(n+d)...(n+kd)}=a_{n+d}-a_n$$
Thus, the sum $\sum_{n=1}^{\infty}\frac{1}{n(n+d)...(n+kd)}$ is a telescoping sum.
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