How to calculate $\sum_{n=1}^{\infty}\frac {n^2} {n!}$ [duplicate]

Summation of n²/n! where n takes the values from 1 to ∞

$$\sum_{n=1}^{\infty}\frac {n^2} {n!}$$

marked as duplicate by Hans Lundmark, Nosrati, user491874, Clement C., Guy FsoneJan 13 '18 at 18:02

$$S=\sum_{n=1}^\infty\dfrac{n^2}{n!}=\sum_{n=1}^\infty\dfrac{n(n-1)+n}{n!}=\sum_{n=1}^\infty\dfrac1{(n-2)!}+\sum_{n=1}^\infty\dfrac1{(n-1)!}$$

Set $n-2=m$ in the first sum and $n-1=r$ in the second,

$$S=\sum_{m=0}^\infty\dfrac1{m!}+\sum_{r=0}^\infty\dfrac1{r!}$$ as $\dfrac1{n!}=0$ for $n<0$

Now $$e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$$

• – lab bhattacharjee Jan 13 '18 at 3:25
• so the final answer is 2e right?? – user50973 Jan 13 '18 at 3:39

Another approach: if we have a power series $f(x)=\sum_{n=0}^\infty a_n x^n$ then $\sum_{n=0}^\infty na_n x^n=xf'(x)$, so this is obtained by applying the "differential operator" $x\frac d{dx}$ to $f(x)$. Therefore $$\left(x\frac d{dx}\right)^kf(x)=\sum_{n=0}^\infty n^k a_n x^n.$$

Here take $k=2$ and $f(x)=e^x$. Then $$\sum_{n=0}^\infty \frac{n^2}{n!}x^n=\left(x\frac d{dx}\right)^2e^x =\left(x\frac d{dx}\right)(xe^x)=x(e^x+xe^x)=(x+x^2)e^x.$$ Setting $x=1$ gives $$\sum_{n=0}^\infty \frac{n^2}{n!}=2e.$$

The advantage of this method is that it applies to all power series, and does not involve any particular ingenuity.

Hint

For $n>1$

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

Or this way...

$$S=\sum_{n=1}^{\infty}\frac {n^2} {n!}=\sum_{n=1}^{\infty}\frac {n} {(n-1)!}=\sum_{n=0}^{\infty}\frac {n+1} {n!}=\sum_{n=0}^{\infty}\frac 1 {n!}+\sum_{n=0}^{\infty}\frac {n} {n!}=\sum_{n=0}^{\infty}\frac 1 {n!}+\sum_{n=1}^{\infty}\frac {1} {(n-1)!}$$

$$S=\sum_{n=0}^{\infty}\frac 1 {n!}+\sum_{n=0}^{\infty}\frac {1} {n!}=2\sum_{n=0}^{\infty}\frac 1 {n!}=2e$$