Interval of Convergence for $\sum_{n=1}^{\infty} \frac{n^2 x^n}{2\cdot4\cdot 6 \cdots 2n}$ 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.
 A: 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.
A: I think you made a mistake in computation..
$ \displaystyle \frac{a_{n+1}}{a_n} $ term does not have $2n$ term in numerator..
so its ratio limit goes 0 for every x, so series converges for all real number x.
for example, if you take x=1, series becomes $ \sum_{n} \frac{n^2}{2^n n!} $
which converges since it is smaller than $ \sum_{n} \frac{n^2}{2^n} $ and this is convergent.
A: I think ratio test is probably not the best option for this. Consider the root test. Then we have $(n^{2}x^{n})^{1/n}=n^{2/n}x$ and $2*4*6...*2n=2^{n/2}*n!$'s $n$'th root is $2^{1/2}*(n!)^{1/n}$. We may disregard the constant $2^{1/2}$, and $n^{-2/n}*(n!)^{1/n}$ should be great than 1. And for fixed $x$ it should be greater than $x$ once $n$ goes very large. Hence the series should converge for any $x$. This is not rigorous but may provide an "intuitive" way to look at it when you do not have pen and paper at hand. 
