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.
Best Answer
You made an algebraic error:
$$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2 x^{n+1}}{(2n+2)n^2x^n}=\frac{(n+1)^2x}{2n^2(n+1)}=\frac{(n+1)x}{2n^2}\;,$$ which converges to $0$ for all $x$.
Note, though, that even without this error what you wrote has a few technical problems. Here’s the first step with the algebra corrected but otherwise just as you wrote it:
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \frac{(n+1)^2 x^{n+1}}{(2n+2)\cdot{n^2}\cdot{x^n}}$$
This is false; $$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2 x^{n+1}}{(2n+2)\cdot{n^2}\cdot{x^n}}$$ is true, and so is $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty}\frac{(n+1)^2 x^{n+1}}{(2n+2)\cdot{n^2}\cdot{x^n}}\;,$$ but not the mixed version that you wrote.
I realize that you probably knew what you meant, but you can’t afford to be that sloppy: not only will you confuse readers, but you’re likely eventually to confuse yourself on occasion as well.