# Find $\left[ \sum \limits^{2006}_{k=1}\left( -1\right) ^{k}\frac{k^{2}- 3}{\left( k+1\right) ! } \right] -1$

$$\left[ \sum \limits^{2006}_{k=1}\left( -1\right) ^{k}\frac{k^{2}- 3}{\left( k+1\right) ! } \right] -1$$ 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

• According to Mathematica the answer is $4.657805200972047\times10^{-5756}!$ (That's no a factorial sign; I'm just surprised.) Sep 17, 2015 at 0:24
• @DirkGently ok. But I asked about the closed form Sep 17, 2015 at 0:25
• 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. Sep 17, 2015 at 0:29

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.

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