$$ \left[ \sum \limits^{2006}_{k=1}\left( -1\right) ^{k}\frac{k^{2}- 3}{\left( k+1\right) ! } \right] -1$$ My question is how I can get this summation in the closed form?. I tried to evaluate the summation . But I couldn't complete. How I can solve it?. Thanks

  • 2
    $\begingroup$ According to Mathematica the answer is $4.657805200972047\times10^{-5756}!$ (That's no a factorial sign; I'm just surprised.) $\endgroup$
    – DirkGently
    Sep 17, 2015 at 0:24
  • $\begingroup$ @DirkGently ok. But I asked about the closed form $\endgroup$ Sep 17, 2015 at 0:25
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    $\begingroup$ That's why this is a comment and not an answer. Please also remember that the comments and answers here are not only for the questioner but for anyone who might find the question interesting. $\endgroup$
    – DirkGently
    Sep 17, 2015 at 0:29

1 Answer 1


The closed form is

$$\frac{(-1)^N (N-1)}{6 (4)_{N-2}}$$

for the upper limit of the sum $N$

i.e. for $N=2006$

$$\frac{(-1)^{2006} 2005}{2007!}$$

Proof we can verify that the difference of this expression with itself with $N\to N-1$ gives the summand $$ -\frac{(-1)^N N}{(N+2)!}-\frac{(-1)^N (N-2)}{(N)!} = \frac{(-1)^N \left(N^2-3\right)}{(N+1)!} $$ and also the expression vanishes for $N=1$. So we get a proof by induction.

  • $\begingroup$ or $\frac{(-1)^N (N-1)}{(N+1)!}$ is a slightly better notations $\endgroup$ Sep 17, 2015 at 0:32
  • $\begingroup$ Thanks for help. But could you please explain How we get it? $\endgroup$ Sep 17, 2015 at 0:34
  • $\begingroup$ need a little explain about. How we get this? . $\endgroup$ Sep 17, 2015 at 0:39
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    $\begingroup$ added a little explanation. Hope this helps! $\endgroup$ Sep 17, 2015 at 0:46
  • $\begingroup$ A more practical explanation: type ${\rm Sum}[(-1)^k (k^2 - 3)/(k + 1)!, k]$ in mathematica $\endgroup$ Sep 17, 2015 at 0:48

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