<s>[INST] How to compute the following series using taylor expansion manipulation? How to compute $\sum^{\infty}_{n=0} \frac{x^n}{(n+2)n!}$ and $\sum^{\infty}_{n=0}(-1)^n \frac{(n+1)x^{2n+1}}{(2n+1)!}$ using taylor expansion manipulation? 
$1.\sum^{\infty}_{n=0} \frac{x^n}{(n+2)n!}=\frac{1}{x^2}\sum^{\infty}_{n=0} \frac{x^{n+2}}{(n+2)n!}=\frac{1}{x^2}\int\sum^{\infty}_{n=0} \frac{x^{n+1}}{n!}=\frac{x}{x^2}\int\sum^{\infty}_{n=0} \frac{x^{n}}{n!}=\frac{x}{x^2}e^x=\frac{e^x}{x}$
$2. \sum^{\infty}_{n=0}(-1)^n \frac{(n+1)x^{2n+1}}{(2n+1)!}=\frac12\sum^{\infty}_{n=0}(-1)^n \frac{(2n+2)x^{2n+1}}{(2n+1)!}=\frac12 \frac{d}{dx}\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+2}}{(2n+1)!}=\frac{x}{2} \frac{d}{dx}\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!}=\frac{x\cos{x}}{2}$
Is my process and result correct?
 [/INST] There were some flaws in the development in the OP.  Following that development, we write instead
$$\begin{align}
\sum^{\infty}_{n=0} \frac{x^n}{(n+2)n!}&=\frac{1}{x^2}\sum^{\infty}_{n=0} \frac{x^{n+2}}{(n+2)n!}\\\\
&=\frac{1}{x^2} \sum^{\infty}_{n=0} \int_0^x\frac{t^{n+1}}{n!}\,dt\\\\
&= \frac{1}{x^2}\int_0^x \sum^{\infty}_{n=0} \frac{t^{n+1}}{n!}\,dt\\\\
&=\frac{1}{x^2}\int_0^x t\sum^{\infty}_{n=0} \frac{t^{n}}{n!}\,dt\\\\
&=\frac{1}{x^2}\int_0^x te^t\,dt\\\\
&=\frac{1}{x^2}\left(e^x(x-1)+1\right)
\end{align}$$

For the second one, we have
$$\begin{align}
\sum^{\infty}_{n=0}(-1)^n \frac{(n+1)x^{2n+1}}{(2n+1)!}&=\frac12\sum^{\infty}_{n=0}(-1)^n \frac{(2n+2)x^{2n+1}}{(2n+1)!}\\\\
&=\frac12 \frac{d}{dx}\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+2}}{(2n+1)!}\\\\
&=\frac12 \frac{d}{dx}\left(x \sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!}\right)\\\\
&=\frac12 \frac{d}{dx}\left(x \sin(x)\right)\\\\
&=\frac12 (x\cos(x)+\sin(x))
\end{align}$$
</s>