Prove that a power series satifies that $-f''(x) + x^2 f(x) = 3f(x)$ for all $x \in \mathbb{R}$ 
Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ of a power series defined as
  $$
f(x) := \sum_{n=0}^\infty \frac{(-1)^n2^{-n}x^{2n+1}}{n!}, \ x \in \mathbb{R}
$$
Question:
Find the power series for $f'(x)$ and $f''(x)$ and prove that 
  $$
-f''(x) + x^2f(x) = 3f(x) \ \text{for all} \ x \in \mathbb{R}
$$

In an earlier question I have argued for that $f \in C^\infty(\mathbb{R})$ as we have a Power Series of the form $\sum_{n=0}^\infty a_n(z-a)^n$ with $a = 0 \in \mathbb{R}$ and convergence radius $R = \infty > 0$ which means that $f \in C^\infty(\mathbb{R})$ in the interval $]-\infty, \infty[$. Thus we have that
$$
f'(x) = \sum_{n=0}^\infty \frac{(2n+1)(-1)^n2^{-n}x^{2n}}{n!} \ \text{for} \ -\infty < x < \infty
$$
and
$$
f''(x) = \sum_{n=0}^\infty \frac{(4n^2+2n)(-1)^n2^{-n}x^{2n-1}}{n!} \ \text{for} \ -\infty < x < \infty
$$
but then to show that 
$$
-f''(x) + x^2f(x) = 3f(x) \ \text{for all} \ x \in \mathbb{R}
$$
is a struggle for me. We have not learned (and I am not even aware if we are going to learn) how to manipulate sums so I hope some of you can help me understand step by step how to solve this differential equation. I got an answer on this question a few days ago but I thought it was too difficult to understand so I hope you do not mind helping me again.
I start of by calculating the LHS:
\begin{align*}
    -f''(x) + x^2f(x) & = - \left( \sum_{n=0}^\infty \frac{(4n^2+2n)(-1)^n2^{-n}x^{2n-1}}{n!} \right) + x^2 \cdot \left( \sum_{n=0}^\infty \frac{(-1)^n2^{-n}x^{2n+1}}{n!} \right) \\
    & = 
\end{align*}
and now I think I have to get the same exponents for $x$ but how would one do this? How do I go from $x^{2n-1}$ to $x^{2n+1}$ by rearraging the sum? I can't see how going from $n = 0$ to $n=1$ as for $n=0$ the sum is just equal to $0$ would change anything. What do I do?
If you mind helping me would you please explain the steps you do if it is not too much to ask for.
Thanks in advance! :) 
 A: $$
  S= -  \sum_{n=0}^\infty \frac{(4n^2+2n)(-1)^n2^{-n}x^{2n-1}}{n!} $$
Note that you have to change the indice of the  sum because it's zero for $n=0$. So we start the indice $n$ at $n=1$.
$$
  S= - \sum_{n=1}^\infty \frac{(4n^2+2n)(-1)^n2^{-n}x^{2n-1}}{n!} $$
Since we have $f$ on RHS which has $x^{2n+1}$ we will try to change our serie so it looks like that. We can change the indice $n$ so that the indice becomes $2m+1$
$$2n-1=2m+1 \implies n=m+1$$
Change the sum now. The series starts at $n=1 \implies m=0$:
$$
  S= - \sum_{m=0}^\infty \frac{(4(m+1)^2+2(m+1))(-1)^{m+1}2^{-1-m}x^{2(m+1)-1}}{(m+1)!} $$
Or more simply:
$$
  S=  \sum_{m=0}^\infty \frac{(2m+3)(-1)^{m}x^{2m+1}}{2^m (m!)} $$
$$S=  \sum_{m=0}^\infty \frac{(2m)(-1)^{m}x^{2m+1}}{2^m (m!)}+ \sum_{m=0}^\infty \frac{3(-1)^{m}x^{2m+1}}{2^m (m!)}$$
And this series is just:
$$  S= 3f(x)+ \sum_{\color{red}{m=1}}^\infty \frac{2m(-1)^{m}x^{2m+1}}{2^m (m!)} $$
$$  S= 3f(x)+ x^2\sum_{{m=1}}^\infty \frac{2m(-1)^{m}x^{2m-1}}{2^m (m!)} $$
We need the series to look like $x^2f(x)$. But this last series starts at zero. We need to change our indice. 
$m=n+1$ Change the series:
$$  S= 3f(x)+ x^2\sum_{{n=0}}^\infty \frac{2(n+1)(-1)^{n+1}x^{2n+1}}{2^{n+1  } ((n+1)!)} $$
Or more simply:
$$  S= 3f(x)- x^2\sum_{{n=0}}^\infty \frac{(-1)^{n}x^{2n+1}}{2^n(n!)} $$
Finally as expected:
$$\boxed {S=3f(x)-x^2f(x)}$$
