Right now I am studying power series and came across a problem in Stewart's Calculus 7th edition that I was unsure of.
I am trying to find the radius of convergence $R$ as well as the interval of convergence $I$ for $\sum_{n=1}^{\infty} \frac{n^2 x^n}{2\cdot4\cdot 6 \cdots 2n}$
$a_n := \frac{n^2 x^n}{2\cdot4\cdot 6 \cdots 2n}$
I began by using Ratio Test
$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \frac{(n+1)^2 x^{n+1} \cdot 2n}{(2n+2)\cdot{n^2}\cdot{x^n}} = \frac{(n^2+2n+1)\cdot x \cdot 2n}{n^2 \cdot (2n+2)} \to |x|$
So, by the Ratio Test, our original series is convergent $\iff$ $|x| \lt 1 \iff -1 \lt x \lt 1 \implies R = 1$.
Then I tried testing for converge at the endpoints $\pm 1$.
For $x=-1$, the power series diverges by the Alternating Series Test, and for $x=1$, the power series diverges by the Divergence Test.
$\therefore I = (-1,1)$.
For some reason, I am having some doubts about the Ratio Test part, but if someone could check my work, that would be nice.