Answer
$\frac{1}{2!}$ + $\frac{2}{3!}$ + $\frac{3}{4!}$ + ... + $\frac{n}{(n+1)!}$
$\sum_{k=0}^N \frac{k+1}{(k+2)!}$
Work Step by Step
The sequence begins from 0 until n .
When k=0, the term (i.e, that particular element of the sequence) is evaluated to $\frac{0+1}{(0+2)!}$ . This is why the term is $\frac{1}{2!}$
When k=1, the term (i.e, that particular element of the sequence) is evaluated to $\frac{1+1}{(1+2)!}$ . This is why the term is $\frac{2}{3!}$
When k = 2, the term (i.e, that particular element of the sequence) is evaluated to $\frac{2+1}{(2+2)!}$. This is why the term is $\frac{3}{4!}$