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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!

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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

4 Answers 4

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, $S=\dfrac1{96}$.

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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
1  
@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

$$\sum_{n=3}^{m}\frac{(n-3)!}{(n+2)!}=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}$$

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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 $\dfrac{1}{96}$.

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

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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

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