Question

Find the sum of the series $\frac{1^2}{1!}+\frac{2^2}{2!}+\frac{3^2}{3!}+....\infty$.

Answer 1

For the given series $\frac{1^2}{1!}+\frac{2^2}{2!}+\frac{3^2}{3!}+....\infty$, the $n^{th}$ term be $\frac{n^2}{n!}=\frac{n}{(n-1)!}=\frac{(n-1)+1}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}$.

Then the given series can be represented as

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

$=(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+....\infty )$$+(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+....\infty )$

$=e+e$

$=2e$.