# Find the sum to infinity of the series $1 + \frac{4}{3!} + \frac{6}{4!} + \frac{8}{5!} + \cdots$

Sum the series to infinity

$$1 + \frac{4}{3!} + \frac{6}{4!} + \frac{8}{5!} + \cdots$$

The general term is:

$$u_n = \frac{2n}{(n+1)!}$$

Not sure how to tackle this series.

## 1 Answer

\begin{align} \sum_{n=1}^\infty\frac{2n}{(n+1)!}&=2\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-2\sum_{n=1}^\infty\frac{1}{(n+1)!}\\ &=2\sum_{n=1}^\infty\frac{1}{n!}-2\sum_{n=1}^\infty\frac{1}{(n+1)!}\\ &=2\sum_{n=1}^\infty\frac{1}{n!}-2\sum_{n=2}^\infty\frac{1}{n!}\\ &=2\cdot\frac1{1!}=2. \end{align}

• You are basically observing that $$\frac{2n}{(n+1)!}= \frac{2}{n!}-\frac{2}{(n+1)!}$$ which makes the series telescopic. Sep 21, 2020 at 22:51