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If the sum $$\sum_{n=0}^{2011} \frac{n+2}{n!+(n+1)!+(n+2)!}$$ can be written as $$\frac{1}{2} - \frac{1}{a!}$$find the last three digits of a.

I have reduced the given expression to $$\frac{1}{(n+2)(n)!}$$ and I think I will have to use the method of differences but I don't know how to proceed. Any help would be appreciated.

graydad
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user140161
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2 Answers2

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$$\begin{align} \sum_{n=0}^{2011} \frac{n+2}{n!+(n+1)!+(n+2)!} &=\sum_{n=0}^{2011}\frac{1}{(n+2)(n)!} \\ &=\sum_{n=0}^{2011}\frac{n+1}{(n+2)!}\\ &=\sum_{n=0}^{2011}\frac 1{(n+1)!}-\frac 1{(n+2)!}\\ &=\sum_{n=0}^{2011}\frac 1{(n+1)!}-\sum_{n=1}^{2012}\frac 1{(n+1)!}\\ &=\frac 1{(0+1)!}-\frac 1{(2012+1)!}\\ &=1-\frac 1{2013!} \end{align}$$

NB - if the sum is taken from $n=1$ instead of $n=0$, then the result becomes $$\frac 12-\frac 1{2013!}$$ which is in the form required, with the last three digits of $a$ being $013$.

Hypergeometricx
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Hint: $$\frac1{(n+k)!}-\frac1{(n+k+1)!}=\frac{n+k}{(n+k+1)!}=\frac1{(n+k+1)(n+k-1)!}\ .$$

David
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  • I understand this result but can you please tell me how you arrived at it? Also, I am having trouble applying this result to my expression – user140161 Dec 21 '14 at 21:25
  • If, as stated in the answer, the sum is $\frac12-\frac1{a!}$, then your expression pretty much has to be something like my LHS. Then all you have to do is find the value of $k$ to match your expression. – David Dec 21 '14 at 22:26