Exercise 385Calculus › Single-variable calculus › Series › Find the sumL1Added 2 Nov 2026

Series with factorials and the number e

#exponential series#factorials#e

Statement

Using $\displaystyle\sum_{n=0}^\infty\frac1{n!}=e$ , compute
$\sum_{n=1}^\infty\frac n{n!},\qquad\sum_{n=0}^\infty\frac{n^2}{n!},\qquad\sum_{n=0}^\infty\frac1{(n+1)!}.$

Hint

Rewrite $n^2=n(n-1)+n$ and use $\frac n{n!}=\frac1{(n-1)!}$ and $\frac{n(n-1)}{n!}=\frac1{(n-2)!}$ .

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Solution

First. For $n\ge1$ , $\frac n{n!}=\frac1{(n-1)!}$ , so the sum is $\sum_{m\ge0}\frac1{m!}=e$ .

Second. Since $n^2=n(n-1)+n$ and the terms $n=0,1$ vanish for $n(n-1)$ ,
$\sum_{n\ge0}\frac{n^2}{n!}=\sum_{n\ge2}\frac1{(n-2)!}+\sum_{n\ge1}\frac1{(n-1)!}=e+e=2e.$

Third. $\sum_{n\ge0}\frac1{(n+1)!}=\sum_{m\ge1}\frac1{m!}=e-1$ .

Series with factorials and the number e

Statement

Solution

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