How to evaluate $ \:\sum _{n=3}^{\infty \:}\frac{4n^2-1}{n!}\:\: $?

Hint: $$ \frac{k^2}{k!} = \frac{k-1+1}{(k-1)!} = \frac{1}{(k-2)!} + \frac{1}{(k-1)!} $$

Note that $$e= \sum _{n=0}^{\infty}\frac{1}{n!}$$ \begin{split} \sum _{n=3}^{\infty }\frac{4n^2-1}{n!} &= 4\sum _{n=3}^{\infty }\frac{n^2}{n!} -\sum _{n=0}^{\infty}\frac{1}{n!}+\frac{5}{2}\\ &=4\sum _{n=3}^{\infty }\frac{n-1+1}{(n-1)!} -\sum _{n=0}^{\infty}\frac{1}{n!}+\frac{5}{2}\\ &= 4\sum _{n=3}^{\infty }\frac{1}{(n-1)!} + 4\sum _{n=3}^{\infty }\frac{1}{(n-2)!} -\sum _{n=0}^{\infty}\frac{1}{n!}+\frac{5}{2}\\ &= \frac{5}{2}+4\sum _{j=2}^{\infty \:}\:\:\frac{1}{j!} + 4\sum _{k=1}^{\infty }\frac{1}{k!} -\sum _{n=0}^{\infty}\frac{1}{n!} \\ &= \frac{5}{2}+4(e-2)+ 4(e-1)-e=7e-\frac{19}{2} \end{split}

For higher powers (series of the form $\displaystyle \sum_{k\geq 0}\frac{k^n}{k!} , n > 0 $ ) we use Dobínski's formula and the our problem is just particular case of it. .

$$4\sum_{n\geq 3}\frac{n^2}{n!}-\sum_{n\geq 3}\frac{1}{n!}=4\left(\sum_{n\geq 0}\frac{n^2}{n!} -1-\frac{2^2}{2!}\right)-\left(\sum_{n\geq 0}\frac{1}{n!}-2-\frac{1}{2!}\right)=4\sum_{n\geq 0}\left(\frac{n^2}{n!}-\frac{1}{n!}\right)-12+\frac{5}{2}$$ and hence by Dobiński's formula we have $$4eB_2-eB_0-\frac{19}{2}= 8e-e-\frac{19}{2}=7e -\frac{19}{2} $$ where $B_n$ is nth Bell number with $B_0=B_1=1$ and $B_2=2$