Given the (complex) power series $\sum_{n=0}^{\infty} \frac{(-1)^n 2^{(-n)}z^{(2n+1)}}{n!}$, I was to find first; the radius of convergence, and then the derivative and second derivative of the (real-valued) sum function $f(x)=\sum_{n=0}^{\infty} \frac{(-1)^n 2{(-n)}x^{(2n+1)}}{n!} ,x\in\mathbb{R}$.
Having found these (and now we get to the question proper) I was to show that $x^2f(x)-f''(x)=3f(x)$. And the last part here rather troubles me.
Radius of convergence is $R=\infty$, where I simply used $R=\lim_{n\rightarrow\infty} \frac{|a_n|}{|a_{n+1}|}$. This particular approach is for series in $\sum... |z-a|^n$, but I think it's justified because I could always substitute $z^{2n+1}=y^n$, for some $y$, right?
The derivatives of the (real) sum function are
\begin{equation} f'(x)=\sum_{n=0}^{\infty} a_n (2n+1) x^{2n+1-1} = \sum_{n=0}^{\infty} a_n (2n+1) x^{2n} \end{equation} & \begin{equation} f''(x)=\sum_{n=1}^{\infty} a_n (2n+1)(2n) x^{2n+1-1-1} = \sum_{n=1}^{\infty} a_n (2n+1)(2n) x^{2n-1} \end{equation}
Assuming I've done this correctly. To show that $x^2f(x)-f''(x)=3f(x)$, I simply started trying to calculate $f''(x)+3f(x)$, which then ought to equal $x^2f(x)$. But I can't seem to get there. My work thus far (with tediously many steps written, to show the process)
\begin{align*} & \sum_{n=1}^{\infty} a_n (2n)(2n+1) x^{2n-1} + 3\sum_{n=0}^{\infty} a_n x^{(2n+1)}\\ =& \sum_{n=0}^{\infty} a_{n+1} (2n+2)(2n+3) x^{2n+1} + 3\sum_{n=0}^{\infty} a_n x^{(2n+1)}\\ =& \sum_{n=0}^{\infty} \frac{(-1)^{n+1}2^{-n-1}}{n!(n+1)} 2(n+1)(2n+3) x^{2n+1} +3\sum_{n=0}^{\infty} \frac{(-1)^{n}2^{-n}}{n!} x^{(2n+1)}\\ =& \sum_{n=0}^{\infty} \frac{(-1)^{n+1}2^{(-n)}}{n!} (2n+3) x^{2n+1} +3\sum_{n=0}^{\infty} \frac{(-1)^n 2^{(-n)}}{n!} x^{(2n+1)}\\ =& \sum_{n=0}^{\infty} -a_n (2n+3) x^{2n+1} + 3a_n x^{(2n+1)}\\ =& \sum_{n=0}^{\infty} a_n x^{(2n+1)}\bigg(-(2n+3)+ 3\bigg)\\ =& \sum_{n=0}^{\infty} a_n x^{(2n+1)}(-2n)\\ \end{align*}
I cannot, however, see how this should ever equal $x^2f(x) = x^2\sum_{n=0}^{\infty} \frac{(-1)^n 2{(-n)}x^{(2n+1)}}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n 2{(-n)}x^{(2n+3)}}{n!}$. The exponents of x just don't add up.
P.s. in the first line in the above set of equations, I shifted $n\rightarrow n+1$ such that the double-derivative function is truncated to start at $n=0$, rather than $n=1$.
With a little help from a fellow student, I figured it out. In case anyone ever has a similar problem, I'll leave the solution here.
Starting with the last line from the above: \begin{align*} =& \sum_{n=0}^{\infty} a_n x^{(2n+1)}(-2n)\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n 2^{(-n)}}{n!} x^{(2n+1)}(-2n)\\ =& x^2\sum_{n=1}^{\infty} \frac{(-1)^{n-1} 2^{(1-n)}}{(n-1)!} x^{(2n-1)}\\ \end{align*} The last sum starts at $n=1$, because it makes no sense for $n=0$ (where we'd get stuff like $(-1)!$; so we shift it to $n=1$). To shift it back down to starting at $n=0$, we have to add +1 to each $n$. \begin{align*} =& x^2\sum_{n=1}^{\infty} \frac{(-1)^{n-1} 2^{(1-n)}}{(n-1)!} x^{(2n-1)}\\ =& x^2\sum_{n=0}^{\infty} \frac{(-1)^{n} 2^{(-n)}}{n!} x^{(2n+1)}\\ =& x^2 f(x) \end{align*}
Which is what was to be shown.