Find the Sum of the Series $\sum_{n=0}^\infty \frac{3n^2 -1}{(n+1)!}$

Question

Find the Sum of the Series $$\sum_{n=0}^\infty \frac{3n^2 -1}{(n+1)!}$$ I separated the Series in to the sum of $\sum_{n=0}^\infty \frac{3n^2}{(n+1)!}$ and $\sum_{n=0}^\infty \frac{-1}{(n+1)!}$. First i proceeded to find the sum of the Series $\sum_{n=0}^\infty \frac{-1}{(n+1)!}$. What i did is to integrate $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$, then $e^x = \sum_{n=0}^\infty \frac{x^n x}{(n+1)n!} = \sum_{n=0}^\infty \frac{x^n x}{(n+1)!}$. Finally i've got that $$\frac{e^x}{x}=\sum_{n=0}^\infty \frac{x^n }{(n+1)!}$$. So $\sum_{n=0}^\infty \frac{-1}{(n+1)!}$ should be equal to $e$ if i choose $x=1$. The problem is that when i calculate the Sum with wolfram alpha https://www.wolframalpha.com/input/?i=sum+%281%29%2F%28%28n%2B1%29%21%29+%2Cn%3D0+to+infinity the resault is other. It seems to be missing a term. The Sum $\sum_{n=0}^\infty \frac{3n^2}{(n+1)!}$ i'm not really sure how to calculate it. Thanks in advance

Answer 1

$$\sum_{n=0}^\infty\frac{1}{(n+1)!}=\sum_{m=1}^\infty\frac1{m!}=e-1,$$ $$\sum_{n=0}^\infty\frac{n+1}{(n+1)!}=\sum_{n=0}^\infty\frac1{n!}=e,$$ $$\sum_{n=0}^\infty\frac{(n+1)n}{(n+1)!}=\sum_{n=1}^\infty\frac1{(n-1)!}=e.$$ If you can express $$\sum_{n=0}^\infty\frac{3n^2-1}{(n+1)!}$$ as a linear combination of these sums, then you're in business.

Answer 2

Actually,$$\sum_{n=0}^\infty\frac1{(n+1)!}=\sum_{n=1}^\infty\frac1{n!}=\left(\sum_{n=0}^\infty\frac1{n!}\right)-1=e-1.$$

And$$\sum_{n=1}^\infty\frac{n^2}{(n+1)!}=e-1$$as was proved here.

Answer 3

We have $$ 3n^2-1=3(n+1)^2-6(n+1)+2 $$ Hence $$ \sum_{n=0}^\infty \frac{3n^2-1}{(n+1)!}= \sum_{n=0}^\infty \frac{3(n+1)^2-6(n+1)+2}{(n+1)!}= \sum_{n=0}^\infty \frac{3(n+1)^2}{(n+1)!}-\sum_{n=0}^\infty \frac{6(n+1)}{(n+1)!}+\sum_{n=0}^\infty \frac{2}{(n+1)!}=\\ \sum_{n=0}^\infty \frac{3(n+1)}{n!}-6\sum_{n=0}^\infty \frac{1}{n!}+2\sum_{n=0}^\infty \frac{1}{(n+1)!}=\sum_{n=0}^\infty \frac{3n}{n!}+3e-6e+2(e-1)=3e+3e-6e+2e-2\\=2e-2 $$

Answer 4

$\sum_{n=0}^\infty \frac{3n^2 -1}{(n+1)!}=\sum_{n=0}^\infty \frac{3(n^2 -1)+2}{(n+1)!}=\sum_{n=0}^\infty \frac{3(n-1)}{n!}+\sum_{n=0}^\infty \frac{2}{(n+1)!}=\sum_{n=1}^\infty \frac{3}{(n-1)!}-\sum_{n=0}^\infty \frac{3}{n!}+\sum_{n=0}^\infty \frac{2}{(n+1)!}=3e-3e+(2e-2)=2e-2$