How do I calculate the sum of $\sum_{n=0}^{\infty}{x^{n}\over n!}\cdot{{x^2\over n+2}}?$

Question

How do I calculate the sum of $$\sum_{n=0}^{\infty}{x^{n+2}\over (n+2)n!}=\sum_{n=0}^{\infty}\color{red}{x^{n}\over n!}\cdot{{x^2\over n+2}}?\tag1$$

$$\sum_{n=0}^{\infty}{x^{n}\over n!}=e^x\tag2$$

Answer 1

Hint:

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

$$=x\cdot\dfrac{x^{n+1}}{(n+1)!}-\dfrac{x^{n+2}}{(n+2)!}$$

$$\implies\sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2) n!}=x\sum_{n=0}^\infty\dfrac{x^{n+1}}{(n+1)!}-\sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2)!}$$

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

Answer 2

If $f(x)=\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)n!}$, then $f'(x)=\sum_{n=0}^\infty\frac{x^{n+1}}{n!}=xe^x$. Therefore, since $f(0)=0$, $f(x)=(x-1)e^x+1$.

Answer 3

$$\sum_{n\geq 0}\frac{x^{n+2}}{(n+2)n!}=\sum_{n\geq 0}\int_{0}^{x}\frac{z^{n+1}}{n!}\,dz = \int_{0}^{x}\sum_{n\geq 0}\frac{z^{n+1}}{n!}\,dz = \int_{0}^{x} z e^z\,dz=\left[(z-1)e^z\right]_{0}^{x}. $$ The exchange of $\sum$ and $\int$ is allowed by absolute convergence.