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 $\lim_\limits{n\to\infty}a_n$ ?

  • $\begingroup$ The ratio test applies to series, not sequences. $\endgroup$ – Tim Raczkowski Aug 19 '15 at 1:18
  • 1
    $\begingroup$ @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! $\endgroup$ – Tim Raczkowski 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

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 $\lim\limits_{n\to +\infty}u_n=\infty$


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


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 $\sqrt{2} e^{-10}\approx 0.00006421$.


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 $\approx 0.0000639384$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.