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

I came across a question which required us to find $\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^5-5n^3+4n}$. I simplified it to $\displaystyle\sum_{n=3}^{\infty}\frac{1}{(n-2)(n-1)n(n+1)(n+2)}$ which simplifies to $\displaystyle\sum_{n=3}^{\infty}\frac{(n-3)!}{(n+2)!}$. I thought it might have something to do with partial fractions, but since I am relatively inexperienced with them I was unable to think of anything useful to do. I tried to check WolframAlpha and it gave $$\sum_{n=3}^{m}\frac{(n-3)!}{(n+2)!}=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}$$ From this it is clear that as $m\rightarrow \infty$ the sum converges to $\frac{1}{96}$, however I have no idea how to get there. Any help would be greatly appreciated!

share|cite|improve this question
Are you trying to get $\dfrac{1}{96}$ as your answer? – Henry Apr 13 '12 at 12:18
use partial fraction. – FiniteA Apr 13 '12 at 12:51
@EmileOkada The last summation should be $\sum_{n=3}^{m}\frac{(n-3)!}{(n+2)!}=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2‌​)}$ - fix that – Kirthi Raman Apr 13 '12 at 12:56
@KVRaman the extra sigma was not there originally. It must have been added later. – E.O. Apr 13 '12 at 13:16
@Asaf: Your modification of the title was allright but the modification of the body of the post, much less so. – Did Apr 13 '12 at 13:23
up vote 5 down vote accepted

Hint: There exists some $c_k$ independent of $n$ such that $$ \frac1{(n-2)(n-1)n(n+1)(n+2)}=\sum_{k=-2}^2\frac{c_k}{n+k}. $$ To find $c_k$, multiply both sides by $n+k$ and evaluate the result at $n=-k$. For example, $$ c_{-2}=\left.\frac1{(n-1)n(n+1)(n+2)}\right|_{n=2}=\frac1{24}. $$ Sanity check: $\sum\limits_{k=-2}^2c_k=0$ and every $c_k$ should be a multiple of $\frac1{24}$ with the sign of $(-1)^k$ and depending only on $|k|$.

Once this is done, note that the value $S$ of the series you are looking for is $$ S=c_{-2}\cdot\left(\frac11+\frac12\right)+c_{-1}\cdot\left(\frac12\right)+c_{1}\cdot\left(-\frac13\right)+c_{2}\cdot\left(-\frac13-\frac14\right), $$ which yields the value you got thanks to W|A, namely, $S=\dfrac1{96}$.

share|cite|improve this answer
how do you get $S=c_{-2}\cdot\left(\frac11+\frac12\right)+c_{-1}\cdot\left(\frac12\right)+c_{1}‌​\cdot\left(-\frac13\right)+c_{2}\cdot\left(-\frac13-\frac14\right)$? – noname1014 Apr 13 '12 at 12:41
@TaoHong洪涛: by looking at presence/absence for $n=3,4$, after which the terms cancel since all are present and $\sum c_k=0$. – bgins Apr 13 '12 at 13:00
The critical point is that we are not rearranging terms of a conditionally convergent series, but just calculating an offset of the first few terms; the $n$th terms of the original series and our series are still identical. – bgins Apr 13 '12 at 13:22
@bgins: This is indeed the idea, unfortunately you erred with the red parts of your comment. – Did Apr 13 '12 at 13:28
@TaoHong洪涛: (corrected, thanks Didier:)$$\eqalign{ S &=& c_{-2}\left(\color{green}{1+\frac12}+\frac13+\cdots\right) \\&+& c_{-1}\left(\color{green}{\frac12}+\frac13+\cdots\right) \\&+& c_{ 0}\left(\frac13+\frac14+\cdots\right) \\&+& c_{ 1}\left(\frac14+\frac15+\cdots\right) \\&+& c_{ 2}\left(\frac15+\frac16+\cdots\right) \\&=& \left(\sum_{n=3}^\infty\frac{c_k}{n}\right) + c_{-2}\left(\color{green}{1+\frac12}\right) + c_{-1}\left(\color{green}{\frac12}\right) - c_{ 1}\left(\color{red}{\frac13}\right) - c_{ 2}\left(\color{red}{\frac13+\frac14}\right)}$$ – bgins Apr 13 '12 at 13:35

If you know partial fractions, this should be $$\frac{(n-3)!}{(n+2)!}=\frac{1}{4n}+\frac{1}{24(n-2)}+\frac{1}{24(n+2)}-\frac{1}{6(n-1)}-\frac{1}{6(n+1)}$$

And you might have to simplify the finite sum to get an expression like


share|cite|improve this answer

The Heaviside Method gives the partial fraction decomposition $$ \begin{align} &\frac{1}{(n-2)(n-1)n(n+1)(n+2)}\\[6pt] &=\frac{1}{24(n-2)}-\frac{1}{6(n-1)}+\frac{1}{4n}-\frac{1}{6(n+1)}+\frac{1}{24(n+2)}\tag{1} \end{align} $$ Notice that $$ \frac{1}{24}-\frac{1}{6}+\frac{1}{4}-\frac{1}{6}+\frac{1}{24}=0\tag{2} $$ Therefore, $$ \begin{align} &\sum_{n=3}^m\left(\frac{1}{24(n-2)}-\frac{1}{6(n-1)}+\frac{1}{4n}-\frac{1}{6(n+1)}+\frac{1}{24(n+2)}\right)\\ &=\frac{1}{24}\sum_{n=1}^{m-2}\frac1n-\frac{1}{6}\sum_{n=2}^{m-1}\frac1n+\frac{1}{4}\sum_{n=3}^{m}\frac1n-\frac{1}{6}\sum_{n=4}^{m+1}\frac1n+\frac{1}{24}\sum_{n=5}^{m+2}\frac1n\tag{$\ast$}\\ &=\frac{1}{24}\left(\frac11+\frac12+\frac13+\frac14\right)\\ &-\frac{1}{6}\left(\frac12+\frac13+\frac14+\frac{1}{m-1}\right)\\ &+\frac{1}{4}\left(\frac13+\frac14+\frac{1}{m-1}+\frac{1}{m}\right)\\ &-\frac{1}{6}\left(\frac14+\frac{1}{m-1}+\frac{1}{m}+\frac{1}{m+1}\right)\\ &+\frac{1}{24}\left(\frac{1}{m-1}+\frac{1}{m}+\frac{1}{m+1}+\frac{1}{m+2}\right)\\ &=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}\tag{3} \end{align} $$ where $\displaystyle\sum_{n=5}^{m-2}\frac1n$ is cancelled out of each summation in $(\ast)$ due to $(2)$.

As $m\to\infty$ the $(3)$ tends to $\dfrac{1}{96}$.

share|cite|improve this answer

(n−3)!/(n+2)! = 1/[(n+2)(n+1)n(n-1)(n-2)] and you can easily solve by using fractional part.

share|cite|improve this answer
Read the original question again... – Thomas Apr 13 '12 at 13:26
Sorry. I didn't see that one. – Prasad G Apr 14 '12 at 4:23

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.