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The problem in the book is to determine the radius of convergence for the given power series, and test for convergence at the boundary points.

$$\sum_{n=1}^\infty \frac {(n!)^2}{(2n)!} z^n \qquad z\in\mathbb C$$

Using the ratio test, it is clear that the radius of convergence is 4, that is, for $|z|<4$ the series converges. My question is how can we determine convergence at the boundary?

My Attempt

Both the root test and ratio test are inconclusive. It would be helpful if I could take the limit of $\tfrac{(n!)^2}{(2n)!}4^n$, but I don't know how to directly compute $\lim_{n\to\infty}\tfrac{(n!)^2}{(2n)!}4^n$.

In a similar problem I was able to use some inequalities related to $n!$ to conclude that the sequence was always greater than some positive constant. My thinking is that, if I can show that $\tfrac{(n!)^2}{(2n)!}4^n>c$, for some positive constant c, then the series cannot converge for any complex $z$ where $|z|=4$ since the limit of the series would not be zero. I tried that approach, and here are the inequalities: $$n^n e^{1-n}<n!<n^{n+1}e^{1-n} \qquad\qquad (2n)^{2n}e^{1-2n}<(2n)!<(2n)^{2n+1}e^{1-2n}$$ Rigging this up yields $$\begin{align}\frac{e^{2-2n} n^{2n} 4^n}{(2n)^{2n+1} e^{1-2n}}&<\frac{(n!)^2 4^n}{(2n)!}<\frac{e^{2-2n} n^{2n+2}4^n}{(2n)^{2n} e^{1-2n}}\\\frac{e}{2n}&<\frac{(n!)^2 4^n}{(2n)!}<en^2\end{align}$$

Obviously, this isn't going to cut it. I would appreciate a hint.

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This looks like a direct consequence of Stirling's approximation. – Did Oct 15 '11 at 5:22
On the other hand... did Apostol make at least a passing mention of Stirling in his book, before this exercise? – J. M. Oct 15 '11 at 5:40
I don't think so. I also ran a search for "Stirling" on the ebook version, and came up with no results. – process91 Oct 15 '11 at 5:42
The inequalities come from an exercise well before this one in the book, in which we showed $(n!)^{1/n}\sim\tfrac n e$. I got the inequalities for $(2n)!$ in a similar manner, which also yield $[(2n)!]^{1/n}\sim\left(\tfrac {2n} e\right)^2$. This is how I know that the root test is inconclusive. – process91 Oct 15 '11 at 5:53
On the other hand... @user1551 settled the question: Stirling may rest in peace, his formula is not needed here. – Did Oct 15 '11 at 6:18
up vote 8 down vote accepted

Let $a_n=\frac {(n!)^2}{(2n)!} z^n$ with $|z|=4$. Then $\left|\frac{a_{n+1}}{a_n}\right| = \frac {(n+1)^2\times4}{(2n+2)(2n+1)} = \frac{2n+2}{2n+1} >1$. Hence $|a_{n+1}|>|a_n|$ and in turn, $a_n$ does not tend to zero and the infinite series diverges.

BTW, (T.M.?) Apostol has written many books. Which one are you referring to?

share|cite|improve this answer
Calculus, Vol. 1. Thanks, I overlooked that because I thought that would only imply absolute divergence. – process91 Oct 15 '11 at 6:10

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