# What is the telescoping series? $\sum_{k=1}^{n} k \cdot k!$

So we learned telescoping series in class and I came across this question in my textbook and I tried to evaluate it, but I don't understand how to do it.

$$\sum\limits_{k=1}^n\left(k \cdot k!\right)$$

$$\sum\limits_{k=1}^n\left(k \cdot k!\right) = (n+1)! - 1$$
Note that $$k \cdot k! =(k+1)! - k!$$ Now telescoping should help you.
• @user2351149 A simple way to get the right expression when you know what the answer is, is as follows. If you want to show $$\sum_{k=1}^n f(k) = S(n)$$ try to see if $f(k) = S(k+1) - S(k)$. In this case, $((n+1)!-1) - (n!-1) = n \cdot n!$. – user17762 May 5 '13 at 4:20