How should I get the result $$\lim_{n \rightarrow \infty} \frac {e^n(2n)!}{(4n)^nn!}=\sqrt2$$ without relying on Stirling's formula or the Central Limit Theorem?
I am totally clueless. Can somebody help me?
Asked
Active
Viewed 425 times
9
-
7I think you would have to do the equivalent of proving Stirling to get the result. Let me put it this way: if you're not to use Stirling or CLT, what are you allowed to use? – Gerry Myerson Jul 11 '18 at 07:04
-
Unfortunately, the only way of analysing the factorial asymptotically is to use Stirling's formula or CLT. You will definitely be clueless in this case. – Sarvesh Ravichandran Iyer Jul 11 '18 at 08:32
2 Answers
14
This is based on the following lemma.
Lemma: If $f\in C^{1}[0,1] $ then $$\lim_{n\to\infty} \sum_{k=1}^{n}f\left(\frac{k}{n}\right)-n\int_{0}^{1}f(x)\,dx=\frac{f(1)-f(0)}{2}$$ (proof available here)
In the above lemma it is sufficient to assume the Riemann integrability of $f'$.
If $a_n$ is the sequence in question and $L$ is the desired limit then we have \begin{align} \log L&=\log \lim_{n\to\infty}a_n\notag\\ &=\lim_{n\to \infty} \log a_n\notag \\ &=\lim_{n\to\infty} n\log\left(\frac{e}{4}\right)+\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)\notag\\ &=\lim_{n\to\infty} \sum_{k=1}^{n}f\left(\frac{k}{n}\right)-n\int_{0}^{1}f(x)\,dx\notag \end{align} where $f(x) =\log(1+x)$. Using the lemma we see that $$\log L=\frac{\log 2}{2}$$ and hence $L=\sqrt{2}$.
Paramanand Singh
- 87,309
-
2
-
-
@BarryCipra: well I established the result on this assumption of integrability of derivative. I think a simpler proof is possible with just integrability of $f$ and the limit can be expressed as $(f(1)-f(0))/2$. – Paramanand Singh Jul 12 '18 at 03:08
-
-
-
how come in the link the LHS of the lemma is -1 times what you have? – BaroqueFreak Jul 13 '18 at 13:39
-
1@BaroqueFreak: because the sum index in the link goes from $k=0$ to $n-1$. Changing this to $k=1$ to $k=n$ changes the sign and you can easily do the calculations for that. The second link deals with the same formula as used here. – Paramanand Singh Jul 13 '18 at 15:01
-
-6
Solution
According to Stirling's formula, $$n!=\sqrt{2\pi n} \left(\frac{n}{e}\right)^n \cdot e^{\frac{\theta}{12n}}~~,(0<\theta<1)$$
Hence, $$\lim_{n \rightarrow \infty} \frac {e^n(2n)!}{(4n)^nn!}=\lim_{n \to \infty}\frac{e^n\sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n} \cdot e^{\frac{\theta_1}{24n}}}{(4n)^n \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \cdot e^{\frac{\theta_2}{12n}}}=\sqrt{2}.$$
mengdie1982
- 13,840
- 1
- 14
- 39
-
2
-
1I know his intention. But I think there exists no better solution for this without Stirling's formula, unless we are not afraid of the complexity. – mengdie1982 Jul 11 '18 at 08:00