<s>[INST] How to differentiate a power series? I have a task where I need to write the power series:
$\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^n$
differentiated 2 times where $x=0$.
This is what I have done so far:
$f'(x) = \sum\limits_{n=1}^\infty n\frac{(-1)^n}{(2n)!}x^{n-1} \\
    f''(x) = \sum\limits_{n=1}^\infty n(n-1)\frac{(-1)^n}{(2n)!}x^{n-2} = \sum\limits_{n=1}^\infty\frac{((-1)^n (n - 1) n x^{n - 2})}{((2 n)!)}$
However I'm not quite sure how to proceed. 
Update:
So I found the solution by replacing x with 0 which meant that the only term that wouldn't be undefined was 2 which resulted in the sollution:
$ f''(0) = \frac{(-1)^2(2-1)2}{(2*2)!} = \frac{2}{4!} = \frac{2}{4*3*2*1} = \frac{1}{4*3} = \frac{1}{12}$
 [/INST] \begin{eqnarray*}
\sum\limits_{n=1}^\infty\frac{(-1)^n n(n - 1)  x^{n - 2}}{(2 n)!} \mid _{x=0}
\end{eqnarray*}
Note that the first term is zero, the third & higher terms are all zero, so only the second term contributes & gives the value $\color{red}{1/12}$.
Another way to figure is the function is $y=\cos(\sqrt{x})$ now differntiate this twice (chain rule and quotient)
\begin{eqnarray*}
\frac{d^2y}{dx^2} =\frac{ \sin(\sqrt{x})-\sqrt{x} \cos(\sqrt{x})}{4 x^{3/2}}
\end{eqnarray*}
 Now use $\sin(\sqrt{x})=\sqrt{x}-x^{3/2}/6 +\cdots$ and $ \sqrt{x} \cos( \sqrt{x} ) = \sqrt{x} -x^{3/2}/2$ and again we get $\color{red}{1/12}$.
</s>