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I am looking at the sequence and I'm trying to see what happens as $n\to\infty$ $$ a_n = \frac{\prod_{1}^{n}(2n-1)}{(2n)^n} = \frac{1\cdot3\cdot5\cdot...\cdot(2n-1)}{(2n)^n} $$ By inspection, I can see that the denominator increases faster than numerator which indicates that the sequence converges, but I'm not sure how to show that mathematically.

I tried using the ratio test to test for convergence but I got stuck on the following last step

$$ \begin{align*} \lim_{n\to\infty}\left| \frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty}\left|\frac{1\cdot3\cdot5\,\cdot\,...\,\cdot\,(2n-1)(2n+1)}{(2(n+1))^n}\cdot \frac{(2n)^n}{1\cdot3\cdot5\,\cdot\,...\,\cdot\,(2n-1)}\right| \\ &= \lim_{n\to\infty}\frac{n^n(2n+1)}{(n+1)^n} \\ &=\exp\left[\lim_{n\to\infty}\ln\left(\frac{n^n(2n+1)}{(n+1)^n}\right)\right]\\ &=\exp\left[\lim_{n\to\infty}n\ln \left(\frac{n}{n+1}\right) + \ln(2n+1)\right] \end{align*} $$ In short, how do I prove that this sequence is convergent, and how can I evaluate $\lim_\limits{n\to\infty}a_n$ ?

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  • $\begingroup$ The ratio test applies to series, not sequences. $\endgroup$ Aug 19 '15 at 1:18
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    $\begingroup$ @TimRaczkowski: The OP might want to first prove that the corresponding series converges so that the sequence converges to 0. :) $\endgroup$
    – Megadeth
    Aug 19 '15 at 1:19
  • $\begingroup$ @Chou Good point! $\endgroup$ Aug 19 '15 at 1:20
  • $\begingroup$ @rgarcio959. I think there is a little mistake in your set up of the ratio test. That exponent in the denominator of the first fraction should be $n+1$, it shouldn't stay $n$ $\endgroup$
    – imranfat
    Aug 19 '15 at 1:26
  • $\begingroup$ Good catch, I guess I didn't make as much progress as I thought. $\endgroup$
    – rgarci0959
    Aug 19 '15 at 1:28
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As Chou said, if you prove that $\sum_{n=1}^\infty a_n$ converges then $\lim\limits_{n\to +\infty}a_n=0$. In fact, $(a_n)_{n\ge 1}$ is a sequence of positive numbers and $\dfrac{a_{n+1}}{a_n}=\dfrac{n^n(2n+1)}{(n+1)^{n+1}}$ (as imranfat said, there's a little error there). Thus:$$\lim_{n\to +\infty}\dfrac{a_{n+1}}{a_n}=\lim_{n\to +\infty}\dfrac{2n+1}{n+1}\left( \dfrac{n}{n+1}\right)^n$$

Since $\lim\limits_{n\to +\infty}\dfrac{2n+1}{n+1}=2$ and $\lim\limits_{n\to +\infty}\left( 1-\dfrac{1}{n}\right)^n=e^{-1}$ then $\lim\limits_{n\to +\infty}\dfrac{a_{n+1}}{a_n}=\dfrac{2}{e}<1$ and so according to d'Alembert's Criterion, $\sum\limits_{n=1}^\infty a_n$ converges.

You can get the limit from the value of the ratio's limit without d'Alembert's Criterion. Using only sequences, one can prove the following theorem:

Let $(u_n)$ be a sequence of positive real numbers such as $\left(\frac{u_{n+1}}{u_n}\right)$ converges to $\ell\in\mathbb{R}$.

$(i)$ If $\ell <1$ then $\lim\limits_{n\to +\infty}u_n=0$

$(ii)$ If $\ell >1$ then $\lim\limits_{n\to +\infty}u_n=\infty$

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We have $$ \frac{\prod_{1}^{n}(2k-1)}{(2n)^{n}} < \frac{2n-1}{(2n)^{2}} \to 0 $$ as $n$ grows.

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I am not sure that this is an answer to the question but, in any manner, it is too long for a comment.

Inserting the even numbers $$\prod_{i=1}^n (2i-1)=1 \cdot 3\cdot 5 \cdots (2n-1)=\frac {1 \cdot 2 \cdot 3 \cdot 4\cdots (2n-1)\cdot(2n)}{2\cdot 4\cdot 6\cdot 8\cdots \cdot(2n-2)\cdot(2n) }=\frac{(2n)!}{2^n n!}$$ This makes $$a_n = \frac{\prod_{i=1}^{n}(2i-1)}{(2n)^n} = \frac{(2n)!}{2^n\,(2n)^n \,n!}$$ Now, using Stirling approximation of the factorial $$m!\approx \sqrt{2 \pi\, m}\,m^m \,e^{-m}$$ after simplifications $$a_n\approx \sqrt{2} e^{-n}$$ which is a very good approximation and makes the problem much easier (I hope).

For illustration purposes $a_{10}=\frac{26189163}{409600000000}\approx 0.00006394$ while $\sqrt{2} e^{-10}\approx 0.00006421$.

Edit

If you use Stirling's formula to two orders, $$m!\approx \sqrt{2 \pi m}\,m^{m}\,e^{-m} \left(1+\frac{1}{12 m}\right) $$ and use the same procedure as above, you should ned with $$a_n\approx \sqrt{2} e^{-n}\frac{ 1+\frac{1}{24 n}}{1+\frac{1}{12 n}}$$ whic, for $n=10$ would give $\approx 0.0000639399$ while the exact value would be $\approx 0.0000639384$

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