6
$\begingroup$

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)...

$\endgroup$
  • $\begingroup$ Hint: In addition to the answers, try the Comparison test. $\endgroup$ – Amzoti Dec 18 '12 at 1:21
10
$\begingroup$

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.

$\endgroup$
10
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

and each bracket leads to a telescopic sum...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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