Let $z\in\mathbb{C}$ and define the power series by: $$\sum_{n=0}^{\infty} \frac{(-1)^nz^{2n+2}}{(2n+1)!}$$
Notice that the sum starts at $n=0$ so the first term is $z^2$
a) show that the series has a radius of convergence $R = \infty$.
Now let $$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+2}}{(2n+1)!} \ , \ x\in\mathbb{R}$$
b) Find the power series of the functions $f'(x)$ and $f''(x)$ and show that $f''(x)+f(x) = 2\cos(x) \ \forall x\in\mathbb{R}$
a) I know that I can find the radius of convergence by using $R = \lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|} $. However, when the series on the form $\sum_{n=0}^{\infty} \frac{(-1)^nz^{2n+2}}{(2n+1)!}$ do I have to perform a index shift of $z^{2n+2}$ or is there another way to calculate the radius of convergence?
b) $$\begin{align}f'(x) = \sum_{n=0}^{\infty} \frac{d}{dx}\left(\frac{(-1)^nx^{2n+2}}{(2n+1)!}\right) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\cdot(2n+2)x^{2n+1}\\=\sum_{n=0}^{\infty} \frac{d}{dx}\left(\frac{(-1)^n(2n+2)x^{2n+1}}{(2n+1)!}\right)\end{align}$$
$$\begin{align}f''(x) = \sum_{n=0}^{\infty} \frac{d}{dx}\left(\frac{(-1)^n(2n+2)x^{2n+1}}{(2n+1)!}\right) &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\cdot(2n+2)\cdot (2n+1)x^{2n}\\&= \sum_{n=0}^{\infty} \frac{d}{dx}\left(\frac{(-1)^n(2n+2)(2n+1)x^{2n}}{(2n+1)!}\right) \end{align}$$
How am I suppose to conclude that $f''(x) + f(x) = 2\cos(x)$?
Consider the power series
$$ \sum_{n=0}^\infty\frac{(-1)^nw^{n+1}}{(2n+1)!} $$ Then, $a_n=\dfrac{(-1)^n}{(2n+1)!}$, and $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0,$ and hence the about power series, for all $w$, and so does is we replace $w$ by $z^2$.