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)...
Help with infinite sum
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0Hint: In addition to the answers, try the Comparison test. – 2012-12-18
3 Answers
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.
HINT: $\dfrac{1}{n(n+2)(n+4)} = \dfrac{1}{8 n}-\dfrac{1}{4(n+2)}+\dfrac{1}{8 (n+4)}.$
Now write the terms as integrals via $ \int^1_0 x^k dx = \dfrac{1}{k+1}$ and interchange integral and summation. You will have a geometric series inside which you can evaluate, and then you can evaluate the remaining integral.
$\dfrac{1}{n(n+2)(n+4)} = \dfrac{1}{8 n}-\dfrac{1}{4(n+2)}+\dfrac{1}{8 (n+4)}= \left( \dfrac{1}{8 n}-\dfrac{1}{8(n+2)} \right)- \left(\dfrac{1}{8(n+2)}-\dfrac{1}{8 (n+4)} \right)$
and each bracket leads to a telescopic sum...