Question

The value of $\sum _{n=1}^{\infty }\frac{(n^2+1)}{(n+2)n!}$ is:

Answer 1

Now, $\frac{(n^2+1)}{(n+2)n!}$

$=\frac{(n^2-4)+5}{(n+2)n!}$

$=\frac{(n-2)}{n!}+\frac{5}{(n+2)n!}$

$=\frac{(n-2)}{n!}+\frac{5(n+1)}{(n+2)!}$

$=\frac{1}{(n-1)!}-\frac{2}{n!}+\frac{5(n+2-1)}{(n+2)!}$

$=\frac{1}{(n-1)!}-\frac{2}{n!}+\frac{5}{(n+1)!}-\frac{5}{(n+2)!}$.

$\therefore \sum _{n=1}^{\infty }$$\frac{(n^2+1)}{(n+2)n!}$

$=\sum _{n=1}^{\infty }$$\frac{1}{(n-1)!}$$-2\sum _{n=1}^{\infty }$ $\frac{2}{n!}$$+5\sum _{n=1}^{\infty }$$\frac{1}{(n+1)!}$$-5\sum _{n=1}^{\infty }$ $\frac{1}{(n+2)!}$

$=e-2(e-1)+5(e-2)-5(e-2-\frac{1}{2})$

$=\frac{9}{2}-e$.