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 1


\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}

  • 5
    $\begingroup$ You are basically observing that $$\frac{2n}{(n+1)!}= \frac{2}{n!}-\frac{2}{(n+1)!}$$ which makes the series telescopic. $\endgroup$
    – N. S.
    Sep 21, 2020 at 22:51

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