# Series with factorial.

Evaluate $$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}$$

My attempt,

I changed it into $$\sum_{n=1}^{99} \frac{n}{(n+1)!}$$

I really don't know how to attempt for it as it consists of factorial. Can anyone give me some hints? Thanks in advance.

• – lab bhattacharjee May 22 '17 at 1:29
• @labbhattacharjee It helps a lot! – Mathxx May 22 '17 at 1:33
• Hint: $n=(n+1)-1$. – Jack D'Aurizio May 22 '17 at 12:41

## 2 Answers

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

This is a classic example of a telescoping series! It'd first be nice to shift the limits of summation to get the factorial of a more comfortable function, and then we can split up the fraction to show how it telescopes:

\begin{align} \sum_{n=1}^{99} \frac{n}{(n+1)!} &= \sum_{n=2}^{100} \frac{n-1}{n!} \\ &= \sum_{n=2}^{100} \frac{1}{(n-1)!} - \frac{1}{n!} \\ &= \frac{1}{(2-1)!} - \frac{1}{100!} \\ &= 1- \frac{1}{100!} \end{align}

As an aside, note that the result of this series converges towards $1$ as $n$ approaches infinity.