Radius of convergence of $\,x^2\big(1+\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}2^{2n-1}x^{2n}\big)$

Question

I want to calculate the radius of convergence for

$$x^2(1+\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}2^{2n-1}x^{2n})$$

We defined the radius of convergence as $R:=\frac{1}{\lim \sup_{n \to\infty}\sqrt[k](|a_n|)}$ for $\sum_{n=0}^{\infty}a_nz^n$

But how can I bring my power series into the form $\sum_{n=0}^{\infty}a_nz^n$ or how else can I prove that it converges everywhere?

It's the power series of $x^2\cos(x)^2$ if that helps.

Thanks!

Answer 1

Set $w=x^2$, then the power series becomes $$ w\left(1+\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^n2^{2n}w^n}{(2n)!}\right) \tag{1} $$ with coefficents $$ a_n=\frac{(-1)^n2^{2n}}{(2n)!} $$ and $$ \frac{a_{n+1}}{a_n}=-\frac{4}{(2n+1)(2n+2)}, $$ and hence $\,\big|\frac{a_{n+1}}{a_n}\big|\to 0$, which implies that $(1)$ has infinite radius of convergence, and hence converges for every $w\in\mathbb R$ (even in $\mathbb C$). Therefore, the original power series also converges for every $w\in\mathbb R$ (even in $\mathbb C$), and thus its radius of convergence is infinite.