Question

1+1.1!+2.2!+3.3!+---------+n.n!=?
Please answer my question

Answer 1

The general term of the series $1+1.1!+2.2!+3.3!+...+n.n!$ is

$a_n=n.n!=(n+1-1).n!=(n+1)!-n!$.

The required sum is

$1+\sum_ {k=1}^n a_n=1+\sum_ {k=1}^n [(n+1)!-n!]\\ 1+\sum_ {k=1}^n a_n=1+2!-1!+3!-2!+4!-3!+...+(n+1)!-n!$

The above series is a telescoping series. You can see that the terms cancel out. The sum is

$1+\sum_ {k=1}^n a_n=(n+1)!$