Let $$ S = \sum_{n=0}^\infty \frac{1}{n!(n+2)} $$
Integrate the Taylor Series of $xe^x$ to show that S = 1
Also, Differentiate the Taylor series of $\frac{e^x - 1}{x}$ to show that S = 1
For the integration one, I got $\int xe^x dx$ = $x^2\sum_{n=0}^\infty \frac{1}{n!(n+2)} x^n$ and pretty much stuck. I can see the S term but i have no idea how to move on from there
1) By integration. Let $$F(t):=\int_0^t xe^x dx=[e^x(x-1)]_0^t=e^t(t-1)+1.$$ On the other hand $$F(t)=\int_0^t xe^x dx=\sum_{n=0}^{+\infty} \int_0^t\frac{x^{n+1}}{n!}dx= \sum_{n=0}^{+\infty} \frac{t^{n+2}}{n!(n+2)}.$$ Hence $S=F(1)=1$.
2) By differentiation. Let $$f(x):=\frac{e^x - 1}{x}=\sum_{n=1}^{+\infty} \frac{x^{n-1}}{n!}$$ then $$f'(x)=\frac{e^x}{x}-\frac{e^x - 1}{x^2}.$$ On the other hand $$f'(x)=\sum_{n=1}^{+\infty} \frac{(x^{n-1})'}{n!}= \sum_{n=1}^{+\infty} \frac{(n-1)x^{n-2}}{n!}= \sum_{n=2}^{+\infty} \frac{x^{n-2}}{(n-2)!n} =\sum_{n=0}^{+\infty} \frac{x^{n}}{n!(n+2)}.$$ Hence $S=f'(1)=e-(e-1)=1$.
3) A telescoping approach. Note that $$\frac{1}{n!(n+2)}=\frac{(n+1)}{(n+2)!}=\frac{(n+2)-1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}.$$ Hence $$\sum_{n=0}^N \frac{1}{n!(n+2)}=\sum_{n=0}^N\left(\frac{1}{(n+1)!}-\frac{1}{(n+2)!}\right)=1-\frac{1}{(N+2)!}\to 1$$ as $N\to +\infty$, and we conclude that $S=1$.