Radius of convergence $\Sigma_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$ is my method right?

Question

I am trying to find the radius of convergence of the power series $$\Sigma_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$$

is it true to mark $x^2=t$ than substitude in the original series and get $\Sigma_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}t^{n}$

so that I am avoiding the split of the sequence for $a_k= \frac{(-1)^n}{(2n)!}$ for $k=2n $ and $ a_k= 0 else $

Answer 1

This is a valid first step, sure.

Just remember that if you find a finite radius of convergence for the series in terms of $t$, you then need to go back to the original $x$. For example, if you find the series converges when $|t|<2$, then you know it converges when $|x|<\sqrt{2}$. (Although that won't be a concern for this particular series.)

Answer 2

You can notice that this is the power series of $\cos{x}$, which the radius of convergence of the expansion is the set of all real numbers, or I would advise you to use the ratio test. You shouldn't let $x^2=t$ as you're just overcomplicating it.

Setting up the ratio test: $$\lim_{n \to \infty} \bigg | \frac{{(-1)}^{n+1} x^{2n+2}}{(2n+2)!} \cdot \frac{(2n)!}{{(-1)}^n x^{2n}} \bigg | < 1$$ $$\lim_{n \to \infty} \bigg | \frac{x^{2}}{(2n+1)(2n+2)} \bigg | < 1$$ Try finishing the rest.