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

Question

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

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.