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Recently got this on a test:

$$\lim_{n\to\infty}\frac{\prod_{k=1}^n(2k-1)}{(2n)^n}$$

Because it's a freshman calculus course, I think we were expected to solve it like a physicist. Taking a look at the first few terms of the series:

$$\{\frac{1}{2},\frac{3}{16},\frac{5}{216},\cdots\}$$And saying "this probably converges at $0$". Because this is all that's been covered in our text so far. I find this really sketchy, considering how carefully we normally tiptoe around infinities. What would be a more robust solution of this problem?

Edit

Sorry, the question in the title was different from the one here. Fixed.

Attempted solution by ratio test

$$\lim_{n\to\infty}\frac{\frac{3\cdot5\cdot7\cdot\cdots\cdot(n-1)\cdot n\cdot(n+1)}{(2n+2)^n(2n+2)}}{\frac{3\cdot5\cdot7\cdot\cdots\cdot(n-1)\cdot n}{(2n)^n}}=\lim_{n\to\infty}\frac{2^nn^n(n+1)}{(2n+2)^{(n+1)}}$$Both the top and the bottom have a largest term of $n^{n+1}$, but at the bottom there is a coefficient of $2^{n+1}$, so the series converges.

Is this good reasoning?

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  • $\begingroup$ You could use the "ratio test". Certainly, if the associated series converges, then the sequence must converge to $0$. $\endgroup$ – Omnomnomnom Nov 20 '13 at 18:42
  • $\begingroup$ By ratio test you mean $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$? $\endgroup$ – user99980 Nov 20 '13 at 18:45
  • $\begingroup$ Exactly. You can show in this case $|\frac{a_{n+1}}{a_n}|\to 0$. $\endgroup$ – Omnomnomnom Nov 20 '13 at 18:47
  • $\begingroup$ @Omnomnomnom I've written up a solution by ratio test. I'm not sure if it's correct. Do you mind checking? $\endgroup$ – user99980 Nov 20 '13 at 19:00
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$$ \frac{\prod_{k=1}^n(2\,k-1)}{(2\,n)^n}=\frac{1}{2\,n}\cdot\frac{3}{2\,n}\cdot\dots\cdot\frac{2\,n-1}{2\,n}<\frac{1}{2\,n}. $$

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  • $\begingroup$ Wow, this solution escaped me. Very good solution. I can't believe I missed it. $\endgroup$ – user99980 Nov 20 '13 at 19:09
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Rewrite the product as

$$\prod_{k=1}^n (2 k-1) = \frac{(2 n)!}{2^n n!}$$

Then use the Stirling approximation

$$n! \sim n^n e^{-n} \sqrt{2 \pi n} \quad (n \to \infty)$$

to deduce that the asymptotic behavior of the ratio, which is

$$\frac{(2 n)!}{2^{2 n} n!\, n^n}$$

behaves as $\sqrt{2} e^{-n}$ as $n \to \infty$. Therefore, the limit is zero.

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