I am not finding any path to solve it. Online calculators are not helping either... The answer is it converges and converges to 0.

I believe it is the same as the result of $$\lim_{x\to\infty}\frac1x\times\frac{(2x)!}{4^x\times(x!)^2}$$

I got the idea above by multiplying $a_n$ by ($\frac{2\cdot4\cdot\ldots\cdot2n}{2\cdot4\cdot\ldots\cdot2n}$)

  • $\begingroup$ I've added the bottom rather than replacing it as I'm unsure this is what you were trying to say $\endgroup$ – Henry Lee Mar 17 at 0:41
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    $\begingroup$ The part you added is correct. It is exactly what I tried to say! Thanks. $\endgroup$ – MTLaurentys Mar 17 at 0:43
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    $\begingroup$ Hint: The numerator $1\times3\times\cdots\times(2n-1)$ is less than the denominator $2\times4\times\cdots\times(2n)$. $\endgroup$ – FredH Mar 17 at 0:50
  • $\begingroup$ You can learn some readily-applicable techniques in this answer. $\endgroup$ – Sangchul Lee Mar 17 at 1:22
  • $\begingroup$ Notice that $\frac{1\cdot3\cdot\dots\cdot\left(2n-1\right)}{2\cdot4\cdot\dots\cdot\left(2n\right)}=\frac{\prod_{i=1}^{n}\left(2i-1\right)}{\prod_{i=1}^{n}\left(2i\right)}=\prod_{i=1}^{n}\left(\frac{2i-1}{2i}\right)\le\left(\frac{2n-1}{2n}\right)^{n}$. What can you say about $\lim_{n\to\infty}\frac{1}{n}\left(\frac{2n-1}{2n}\right)^{n}$? $\endgroup$ – Jake Mar 17 at 1:58

I will rewrite $a_{n}$ for clarity. $$a_{n}=\frac{1}{n}\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\cdot\dots\cdot\left(\frac{2n-1}{2n}\right).$$

Notice that the biggest term in this product is $\frac{2n-1}{2n}$ (except for when $n=1$). This means that if we multiply $\frac{2n-1}{2n}$ by itself $n$ times, then $$0\le a_{n}=\frac{1}{n}\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\cdot\dots\cdot\left(\frac{2n-1}{2n}\right)\le b_{n}=\frac{1}{n}\left(\frac{2n-1}{2n}\right)^{n}.$$

Taking the limit of the right side, we have $$\lim_{n\to\infty}b_{n}=\lim_{n\to\infty}\frac{1}{n}\left(\frac{2n-1}{2n}\right)^{n}=\lim_{n\to\infty}\frac{1}{n}\cdot\lim_{n\to\infty}\left(\frac{2n-1}{2n}\right)^{n}.$$

To evaluate $\lim_{n\to\infty}\left(\frac{2n-1}{2n}\right)^{n}$, notice that it is equivalent to $$\lim_{n\to\infty}e^{\ln\left(\left(\frac{2n-1}{2n}\right)^{n}\right)}=\lim_{n\to\infty}e^{n\ln\left(1-\frac{1}{2n}\right)}=\lim_{n\to\infty}e^{\frac{\ln\left(1-\frac{1}{2n}\right)}{1/n}}=e^{\lim_{n\to\infty}\frac{\ln\left(1-\frac{1}{2n}\right)}{1/n}}$$

Using L'Hospital's rule, we get $$e^{\lim_{n\to\infty}\frac{1/\left(2n^{2}\right)}{-1/n^{2}}}=e^{\lim_{n\to\infty}-\frac{1}{2}}=\frac{1}{\sqrt{e}}.$$

Thus, $$\lim_{n\to\infty}\frac{1}{n}\left(\frac{2n-1}{2n}\right)^{n}=\lim_{n\to\infty}\frac{1}{n}\frac{1}{\sqrt{e}}=0.$$

But remember that this is our upper-bound for $a_{n}$, and so by the Squeeze theorem, $$\lim_{n\to\infty}0\le\lim_{n\to\infty}a_{n}\le\lim_{n\to\infty}b_{n}\implies0\le\lim_{n\to\infty}a_{n}\le0.$$

Thus, the limit is $0$.

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    $\begingroup$ L’Hopital’s isn’t really necessary, just rewrite: $$\lim_{n\to\infty} \left(\frac{2n-1}{2n}\right)^n = \lim_{n\to\infty}\left(1+\frac{-1/2}{n}\right)^n = e^{-1/ 2}$$ $\endgroup$ – adfriedman Mar 17 at 18:43
  • $\begingroup$ @adfriedman I considered it but wasn't sure if that would have been familiar or not to the OP. $\endgroup$ – Jake Mar 19 at 0:14

Simple Answer

The product on the right of the "$\cdot$" is bounded above by $1$, so we get $$ \lim_{n\to\infty}\frac1n\,\overbrace{\prod_{k=1}^n\frac{2k-1}{2k}}^{\le1}=0\tag1 $$

Better Bounds

The fact that the product is less than $1$ is enough to prove that the expression tends to $0$ as $n\to\infty$. However, we can get a much better bound using the inequalities $$ \sqrt{\frac{2k-2}{2k}}\le\frac{2k-1}{2k}\le\sqrt{\frac{2k-1}{2k+1}}\tag2 $$ which can be proven by squaring and cross-multiplying. Using $(2)$, we get $$ \frac1{2\sqrt{n}}=\frac12\prod_{k=2}^n\sqrt{\frac{2k-2}{2k}}\le\prod_{k=1}^n\frac{2k-1}{2k}\le\prod_{k=1}^n\sqrt{\frac{2k-1}{2k+1}}=\frac1{\sqrt{2n+1}}\tag3 $$ $(3)$ suggests a more interesting question: find the value of $$ \frac12\le\lim_{n\to\infty}\sqrt{n}\,\prod_{k=1}^n\frac{2k-1}{2k}\le\frac1{\sqrt2}\tag4 $$

Series Rather than Sequence

Perhaps the question was to compute $\sum\limits_{n=1}^\infty a_n$? That sum converges and has a simple closed form.


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} a_{n} & \equiv{1 \over n}\, {1\cdot3\cdot\ldots\cdot\pars{2n - 1} \over 2\cdot4\cdot\ldots\cdot\pars{2n}} = {1 \over n}\,{\prod_{k = 1}^{n}\pars{2k - 1} \over \prod_{q = 1}^{n}\pars{2q}} \\[5mm] & == {1 \over n}\,{2^{n}\prod_{k = 1}^{n}\pars{k - 1/2} \over 2^{n}\prod_{q = 1}^{n}q} = {1 \over n}\,{\pars{1/2}^{\overline{n}} \over n!} = {1 \over n}\,{\Gamma\pars{1/2 + n}/\Gamma\pars{1/2} \over n!} \\[5mm] & = {1 \over n}\, {\pars{n - 1/2}! \over n!}\,{1 \over \root{\pi}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {\pi^{-1/2} \over n}\, {\root{2\pi}\pars{n - 1/2}^{n}\expo{-\pars{n - 1/2}} \over \root{2\pi}n^{n + 1/2}\expo{-n}} \\[5mm] & = {\pi^{-1/2} \over n}\, {n^{n}\bracks{1 - \pars{1/2}/n}^{n}\expo{1/2} \over n^{n + 1/2}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {1 \over \root{\pi}}\,{1 \over n^{3/2}} \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\Large\to}\,\,\,\bbx{0} \end{align}

Note that $\ds{\Gamma\pars{1/2} = \root{\pi}}$.


If $a_n = \frac{1}{n} \cdot \frac{1\cdot3\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot\ldots\cdot(2n)} = \frac{\prod_{k=1}^n(2k-1)}{n\prod_{k=1}^n(2k)} $ then $a_{n+1} = \frac{\prod_{k=1}^{n+1}(2k-1)}{(n+1)\prod_{k=1}^{n+1}(2k)} $ so

$\begin{array}\\ \frac{a_n}{a_{n+1}} &=\frac{(n+1)2(n+1)}{n(2n+1)}\\ &=\frac{2(n^2+2n+1)}{2n^2+n}\\ &=\frac{n^2+2n+1}{n^2+n/2}\\ &=\frac{n^2+n/2+(3/2)n+1}{n^2+n/2}\\ &=1+\frac{(3/2)n+1}{n(n+1/2)}\\ &=1+\frac{(3/2)(n+1/2)+1/4}{n(n+1/2)}\\ &=1+\frac{3}{2n}+\frac{1}{4n(n+1/2)}\\ &>1+\frac{3}{2n}\\ \end{array} $

Since, if $c_k > 0$, then $\prod_{k=1}^n (1+c_k) \ge 1+\sum_{k=1}^n c_k $ (easily proved by induction),

$\begin{array}\\ \frac{a_1}{a_{m+1}} &=\prod_{n=1}^m\frac{a_n}{a_{n+1}}\\ &>\prod_{n=1}^m(1+\frac{3}{2n})\\ &\gt 1+\sum_{n=1}^m\frac{3}{2n}\\ &\gt \frac32\sum_{n=1}^m\frac1{n}\\ &\gt \frac32\ln(m) \qquad\text{(well-known harmonic sum)}\\ \end{array} $

so $a_{m+1} \lt \frac{2a_1}{3\ln(m)} = \frac{1}{3\ln(m)} \to 0$ as $m \to \infty$.

  • $\begingroup$ isn't enough @FredH"s comment? $\endgroup$ – Minz Mar 17 at 2:31
  • $\begingroup$ I wanted to do as much as possible in an elementary way. $\endgroup$ – marty cohen Mar 17 at 2:46
  • $\begingroup$ Is inequality $0<a_n<\frac1n$ not elementary enough? $\endgroup$ – Minz Mar 17 at 2:59

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