# 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.

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Also, if someone knows a good way to deal with the LaTeX for the absolute values, I'd like to know (such as for the ratio $\frac{a_{n+1}}{a_{n}}$ here.) When I tried, it didn't extend over the whole two lines for the fraction and it looked poor. –  Joe Nov 7 '12 at 3:48

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2 x^{n+1}}{(2n+2)n^2x^n}=\frac{(n+1)^2x}{2n^2(n+1)}=\frac{(n+1)x}{2n^2}\;,$$ 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.

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I knew the absolute value was needed (I was not trying to be sloppy), see my comment. I couldn't get them to format nicely. Thanks for the catch on the algebra error - appreciate it. –  Joe Nov 7 '12 at 4:05
@Joe: Sorry: I didn’t notice the comment. If you right-click on any formula and select Display Math As and then TeX Commands, you can see the code that was used; in this case it was just \left| and \right|. You might find this useful if you’ve not already seen it. –  Brian M. Scott Nov 7 '12 at 4:10
Ah, I recall trying that at one point many months ago and it not formatting correctly. I'll use that from now on. Good looks on the Meta post, don't think I've seen that yet! Thanks. –  Joe Nov 7 '12 at 4:12
@Joe: My pleasure! –  Brian M. Scott Nov 7 '12 at 4:15

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.

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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.

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