Q:

$\sum_{k=1}^n \frac{1}{k!(k+1)}$ sigma sum to equation

$$\sum_{k=1}^n \frac{1}{k!(k+1)}$$
I tried for this,
$$\sum_{k=1}^n \frac{k+1}{(k+2)!}$$
I don't know what to do anymore.

A:

If you've taken calculus, you'd know that $\sum_{k=0}^∞ {x^k \over k!} = e^x$.
So, summing $\frac1{k!}\to∞ = e$. Problem is, this includes 1+1... before the actual equation you have. So...
Final answer: $e-2$

