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Wonder how to determine this limit by the use of Riemann integral. The limit is as follows:

$$\lim_{n \to +\infty} \frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}} $$

My instructor told me that the usage of Riemann integral gives spectacular result. Checked Rudin, but did not find any valuable references.

I am very interested in seeing how this "spectacular result" emanates.

Help/advices/solutions very, very appreciated!

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    $\begingroup$ another approach based on sequences is to study the sequence $a_{n} = \dfrac{(2n)!}{n^{n}\cdot n!}$ and calculate the limit of $a_{n + 1}/a_{n}$ as $n \to \infty$. You will very easily see that $a_{n + 1}/a_{n} \to 4/e$ and hence by a standard theorem on limit of sequences $a_{n}^{1/n}$ (and this is the sequence in your question) also tends to $4/e$. $\endgroup$ – Paramanand Singh Jun 3 '16 at 8:00
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Hint: take the logarithm, and try to make something like $\frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)$ appear for $f\colon x\in[0,1] \mapsto \ln (1+x)$. You will find the limit $\ell$ of the logarithm of your quantity, and then by continuity of $\exp$ your answer will be $e^\ell$.


Details. Taking the logarithm,

$$\begin{align} \ln \sqrt[n]{\frac{(2n)!}{n!}} &= \frac{1}{n} \ln \frac{\prod_{k=1}^{2n} k}{\prod_{k=1}^n k} = \frac{1}{n} \ln \prod_{k=n+1}^{2n} k = \frac{1}{n} \sum_{k=n+1}^{2n} \ln k\\ &= \frac{1}{n} \sum_{k=1}^{n} \ln (k+n) = \frac{1}{n} \sum_{k=1}^{n} \ln \left(1+\frac{k}{n}\right) + n\frac{\ln n}{n} \end{align}$$ so that the logarithm of your original quantity is $$\begin{align} \ln \frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}} &= - \ln n + \frac{1}{n} \sum_{k=1}^{n} \ln \left(1+\frac{k}{n}\right) + n\frac{\ln n}{n} = \frac{1}{n} \sum_{k=1}^{n} \ln \left(1+\frac{k}{n}\right) \end{align}$$

Can you see how to use Riemann sums now?

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  • $\begingroup$ @Dr.MV Thanks :) [PS: to the OP, the eventual answer will be $\frac{4}{e} = \exp \int_{0}^1 \ln(1+x) dx$.) $\endgroup$ – Clement C. Jun 2 '16 at 17:51
  • $\begingroup$ I agree with Dr. MV ! Congratulation thousand times !! it's very nice and elegant ! I'm jealous :-D $\endgroup$ – Surb Jun 2 '16 at 19:07
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Applying Stirling's formula, $n! \displaystyle\operatorname*{\sim}_{n\to\infty} \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$, I got the result $\frac{4}{e}$.

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I will repeat basically the same approach as in this answer: Find the value of : $\lim\limits_{n\to \infty} \sqrt [n]{\frac{(3n)!}{n!(2n+1)!}} $ It is also the same approach as suggested in Paramanand Singh's comment. (I see that the OP asks specifically about a proof using Riemann's integral, but this seems interesting enough to be mentioned, too.)

We will use this fact (see the linked answer for references):

Let $(a_n)$ be a sequence of positive real numbers. If $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=L$, then $\sqrt[n]{a_n}$ converges too and $\lim\limits_{n\to\infty}\sqrt[n]{a_n}=L$.

We use the above for $a_n = \frac{(2n!)}{n!n^n}$ and we get $$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim\limits_{n\to\infty} \frac{(2n+1)(2n+2)}{(n+1)^2}\cdot\left(\frac{n}{n+1}\right)^n = \frac4e.$$

This implies that also $$\lim\limits_{n\to\infty} \frac1n\sqrt[n]{\frac{(2n)!}{n!}} = \lim\limits_{n\to\infty} \sqrt[n]{a_n} = \frac4e.$$

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  • $\begingroup$ +1 was lazy enough to write a full answer and therefore posted a comment. And yes this is interesting and perhaps a simpler approach. $\endgroup$ – Paramanand Singh Jun 3 '16 at 10:18

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