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I am trying to show $$\lim_{n \rightarrow \infty} \frac{(2n)!}{n^{2n}}$$

I tried breaking it down, and got stuck when trying to $\left( \frac{2^{n}n!}{n^{n}} \right)$ goes to 0.

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Hint, I think, use Stirling's approximation formula.'s_approximation – Salech Alhasov Feb 16 '13 at 2:10
Use the following result. – Mhenni Benghorbal Feb 16 '13 at 2:17
There is more then one correct and well explained answer below, so I do not know which one to "accept". Thank you all for your brilliant insights! – Jon Feb 16 '13 at 3:03
up vote 2 down vote accepted

By AM-GM we have

$(2n-2)! =2 \cdot 3 \cdot ... \cdot (2n-2) \leq (\frac{2+3+..+2n-2}{2n-3})^{2n-3}=n^{2n-3}$


$$0 \leq \frac{(2n)!}{n^{2n}} \leq \frac{n^{2n-3}(2n-1)(2n)}{n^{2n}}=\frac{(2n-1)(2n)}{n^{3}}$$

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Let $x_n=\dfrac{(2n)!}{n^{2n}}$. Then

$$\begin{align*} \frac{x_{n+1}}{x_n}&=\frac{(2\big(n+1)\big)!}{(n+1)^{2(n+1)}}\cdot\frac{n^{2n}}{(2n)!}\\\\ &=\frac{2(2n+1)}{(n+1)}\left(\frac{n}{n+1}\right)^{2n}\\\\ &=\frac{2(2n+1)}{(n+1)}\left(1-\frac1{n+1}\right)^{2(n+1)-2}\\\\ &=\frac{2(2n+1)}{(n+1)}\left(1-\frac1{n+1}\right)^{2(n+1)}\left(\frac{n+1}n\right)^2\\\\ &=\frac{2(n+1)(2n+1)}{n^2}\left(1-\frac1{n+1}\right)^{2(n+1)}\;. \end{align*}$$



and therefore $\lim\limits_{n\to\infty}x_n=0$.

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Alternative (guidnece for) solution:

Consider $\sum\left( \frac{(2n)!}{n^{2n}} \right)$, that sum is converges, one can use the ratio test to show that, hence the $\left( \frac{(2n)!}{n^{2n}} \right)\to 0$ (Why?)

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true, but I think ratio test depends on sequence going to 0 not the other way around. Would like to see double sided implication justified. – Arjang Feb 16 '13 at 2:24
@Arjang If $\lim \frac{a_{n+1}}{a_n} \in (-1,1)$ it is easy to prove that $\lim_a_n=0$... no series is actually needed. – N. S. Feb 16 '13 at 2:29
@N.S. : Very good point. Why carry the kitchen sink if a drop of water is enough! – Arjang Feb 16 '13 at 3:01

It is not true that $\left( \frac{(2n)!}{n^{2n}} \right)=\left( \frac{2^{n}n!}{n^{n}} \right)$ The left side has a factor $2n-1$ in the numerator while the right side does not. But you can use Stirling's approximation to say $$\frac {(2n)!}{n^{2n}}\approx \frac {(2n)^{2n}}{(ne)^{2n}}\sqrt{4 \pi n}$$ and the powers of $\frac 2e$ take it to zero.

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I very much doubt OP has Stirling's approximation in their arsenal... – vonbrand Feb 16 '13 at 3:07

You definitely can't write $(2 n)!$ as $2^n n!$.

The following bound is easy to prove: $n! \le n^n$ (consider $n! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot n \le n \cdot n \cdot \ldots \cdot n = n^n$), but that isn't enough for you.

A sharper bound can be obtained by considering: $$ (n!)^2 = (1 \cdot n) \cdot (2 \cdot (n - 1)) \cdot \ldots \cdot (n \cdot 1) $$ We now want a bound for the factors. Take the function: $$ f(k) = k (n + 1 - k) $$ To get the maximum we set $f'(k) = 0$, i.e., $n + 1 - 2k = 0$, which gives $k^* = \frac{n + 1}{2}$ As $f''(k^*) = - 2 < 0$, this is the maximum. So: $$ \begin{align*} (n!)^2 &\le \left( \frac{n + 1}{2} \right)^{2 n} \\ n! &\le \left( \frac{n + 1}{2} \right)^n \end{align*} $$ This should make the rest of the proof simple.

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We can write $$\frac{(2 n)!}{n^{2 n}}= \frac{(2 n)(2 n - 1) \ldots (n + 1)}{n^n}\cdot\frac{n!}{n^n}=x_n y_n\;,$$ where $$x_n := \frac{(2 n)(2 n - 1) \ldots (n + 1)}{n^n} = \frac{n + n}{n}\cdots\frac{n + 1}{n} = \left(1 + \frac{n}{n}\right)\ldots\left(1 + \frac{1}{n}\right)$$ and $$y_n := \frac{n!}{n^n}$$ for each $n=1, 2, 3, \ldots$.

Now $x_n$ goes to $1$ as $n$ goes to infinity, whereas the ratio $\frac{y_{n + 1}}{y_n} = \frac{1}{(1 + \frac{1}{n})^n}$ goes to $e^{-1}<1$, which implies that $y_n$ is a monotonically decreasing sequence of positive real numbers and so converges to some non-negative quantity.

Hence the original sequence does converge.

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No, $x_n$ doesn't go to 1 as $n \rightarrow \infty$, in fact, it is larger than 2 always (first factor; others are all ${} > 1$) – vonbrand Feb 16 '13 at 3:06

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