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I was asked to evaluate the following expression:

$\lim_{n \to \infty} \left({\frac{(n+1)(n+2)(n+3)...(3n)}{n{^{2n}}}}\right)^{1/n}$

My first step was to assume that the limit existed, and set that value to $y$.

$ y = \lim_{n \to \infty} \left({\frac{(n+1)(n+2)(n+3)...(3n)}{n{^{2n}}}}\right)^{1/n}$

And then, I took the natural logarithm of both sides of the equation. I obtained the expression:

$ \ln y = \lim_{n \to \infty} \frac{1}{n} \cdot \left(\ln(1+\frac{1}{n}) + \ln(1+\frac{2}{n}) + ... + \ln(1+\frac{2n}{n})\right) $

This simplified to:

$ \ln y = \lim_{n \to \infty} \frac{1}{n} \cdot \sum_{k = 1}^{\color{Red}{2n}} \ln(1+\frac{k}{n}) $

I realize that this is similar to the form of a Riemann sum, which can then be manipulated to give the expression in the form of a definite integral. However, the part bolded in red, which is $ 2n$, throws me off. I have only seen Riemann sums be evaluated when the upper limit is $ n - k $, where $k$ is a constant.

Therefore, how would I go about evaluating this expression?

Thank you for all help in advance.

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  • $\begingroup$ Hint: Use Cesaro-Stolz $\endgroup$ – Paramanand Singh Jul 25 '18 at 8:34
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Consider $$\int_0^2f(x)\,dx$$ where $$f(x)=\ln(1+x).$$ Splitting $[1,2]$ into $2n$ intervals of length $1/n$ gives a Riemann sum $$\frac1n\sum_{k=1}^{2n}f(k/n)=\frac1n\sum_{k=1}^{2n}\ln\left( 1+\frac kn\right)$$ which is exactly yours.

Alternatively you could use Stirling's formula.

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We use the following result: if $(a_n)$ is a sequence such that $a_n >0$ for all $n$ and $(\frac{a_{n+1}}{a_n})$ is convergent, then $(a_n^{1/n})$ is convergent and

$$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=\lim_{n \to \infty}a_n^{1/n}.$$

Now let $a_n:= \frac{(n+1)(n+2)(n+3)...(3n)}{n{^{2n}}}$.

Some easy computations give

$$\frac{a_{n+1}}{a_n}=\frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} \cdot (1- \frac{1}{n+1})^{2n} \to \frac{27}{e^2}.$$

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$(2)^{1/n}\rightarrow 1$, so the answer is not altered by adding an $n$ and dropping $2n$ from the middle of the numerator. Rearrange the terms in the numerator to produce the following pairing: $$((n)(3n))((n+1)(3n-1))((n+2)(3n-2))\cdots ((2n-1)(2n+1))$$

This is $\prod\limits_{i=1}^{n} (2n-i)(2n+i) = \prod\limits_{i=1}^{n} (4n^2-i^2)$ After dividing by $n^{2n}$, you obtain $\prod\limits_{i=1}^{n} (4-(i/n)^2)$. So the limit you are looking for equals to $\exp(\lim\limits_{n\rightarrow \infty} \sum\limits_{i=1}^{n}\frac{1}{n}\log(4-\left(\frac{i}{n}\right)^2))$.

Clearly, as $n$ tends to infinity, the sum $\sum\limits_{i=1}^{n}\frac{1}{n}\log(4-\left(\frac{i}{n}\right)^2))$ tends to the integral $\int\limits_{0}^{1} \log(4-x^2) \, dx = \int\limits_{0}^{1} \log(2-x) \, dx + \int\limits_{0}^{1} \log(2+x) \, dx$.

It is easy to compute both definit integrals.

But your calculation is also good. I that case, you obtain a definite integral on $[0,2]$.

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Just a comment after Lord Shark the Unknown's answer.

When I see things looking like factorials, my first reaction is to think about Stirling's approximation of them.

$$a_n=\left(\frac{\prod_{i=1}^{2n} (n+i)}{n^{2n}}\right)^{\frac 1n}=\left(\frac{(3n)!}{n^{2n}\,n!}\right)^{\frac 1n}$$

$$\log(a_n)=\frac 1n \left(\log((3n)!)-\log(n!)-2n \log(n)\right)$$ Now, Stirling's approximation $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({p}\right)\right)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$ Apply it and simplify to get $$\log(a_n)=\frac 1n \left( (3\log (3)-2)n+\frac{\log (3)}{2}-\frac{1}{18 n}+O\left(\frac{1}{n^3}\right) \right)$$ $$\log(a_n)=(3\log (3)-2)+\frac{\log (3)}{2n}-\frac{1}{18 n^2}+O\left(\frac{1}{n^4}\right) $$ Now, using $$a_n=e^{\log(a_n)}=\frac{27 }{ e^2} \left(1+\frac{\log (3)}{2 n}\right)+O\left(\frac{1}{n^2}\right) $$ which shows the limit and how it is approached.

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