Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)...

share|cite|improve this question
Hint: In addition to the answers, try the Comparison test. – Amzoti Dec 18 '12 at 1:21
up vote 10 down vote accepted



In general:


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.

share|cite|improve this answer
(+1) Nice answer. – Mhenni Benghorbal Jan 2 '13 at 2:43

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.

share|cite|improve this answer
(+1) Nice answer. – Mhenni Benghorbal Jan 2 '13 at 2:44

$$\dfrac{1}{n(n+2)(n+4)} = \dfrac{1}{8 n}-\dfrac{1}{4(n+2)}+\dfrac{1}{8 (n+4)}= \left( \dfrac{1}{8 n}-\dfrac{1}{8(n+2)} \right)- \left(\dfrac{1}{8(n+2)}-\dfrac{1}{8 (n+4)} \right)$$

and each bracket leads to a telescopic sum...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.