evaluate: $ \lim_{n→∞} \frac1n ((n+1)(n+2)(n+3)⋯(2n))^{\frac1n}$.

Question

we have $$ ((n+1)(n+2)(n+3)⋯(2n))=\frac{(2n)!}{n!} $$ Using Stirling formula https://en.wikipedia.org/wiki/Stirling%27s_approximation $$ \log_e(n!)≈n\log_en−n $$ in $$ y=\frac1n((2n)!/n!)^{1/n} $$ we have $$ y=\frac1n\left(\frac{(2n)^{2n} e ^{−2n}}{n^n e^{−n}}\right)^{1/n}=\frac4e $$ So the answer is $$ \lim_{n→∞} \frac1n ((n+1)(n+2)(n+3)⋯(2n))^{\frac1n}=\frac4e $$ Here the question is: $$\lim_{n\to\infty}\frac{1}{n}\Big((n+1)(n+2)\cdots(2n)\Big)^{\frac 1n}~?$$

Answer 1

$$L=\lim_{n\rightarrow \infty}\frac{1}{n} \left ((n+1) (n+2)(n+3)....(2n) \right)^{1/n}$$ $$L=\lim_{n \rightarrow \infty} \left( (1+1/n) (1+2/n) (1+3/n) (1+4/n).....\right)^{1/n}$$ $$\ln L= \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\,\ln\left(1+\frac{k}{n} \right)= \int_{0}^{1} \ln (1+x) dx=(1+x)\ln (1+x)-(1+x)|_{0}^{1}=\ln (4/e)$$ $$\Rightarrow L= 4/e. $$

Answer 2

To get even a bit more than the limit.

$$y=\frac 1 n \frac{(2 n)!}{n!}\implies \log(y)=\log((2n)!)-\log(n!)-\log(n)$$ Using one more term in Stirling approximation and simplifying $$\log(y)=(2 \log (2)-1)+\frac{\log (2)}{2 n}+O\left(\frac{1}{n^2}\right)$$ Continuing with Taylor series $$y=e^{\log(y)}=\frac{4}{e}+\frac{2 \log (2)}{e n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.

Using your pocket calculator, try for $n=10$ (far away from $\infty$ !). The exact result would be $1.52278$ while the above truncated series would give $1.52252$.

Answer 3

If

$\begin{array}\\ f(n) &=((n+1)(n+2)(n+3)⋯(2n))\\ &=\prod_{k=1}^n(n+k)\\ g(n) &=\ln(f(n))\\ &=\sum_{k=1}^n\ln(n+k)\\ &=\sum_{k=1}^n(\ln(n)+\ln(1+k/n)\\ &=n\ln(n)+\sum_{k=1}^n\ln(1+k/n)\\ \text{so}\\ \dfrac{g(n)}{n} &=\ln(n)+\frac1{n}\sum_{k=1}^n\ln(1+k/n)\\ \text{or}\\ \dfrac{g(n)}{n}-\ln(n) &=\frac1{n}\sum_{k=1}^n\ln(1+k/n)\\ &\to\int_0^1 \ln(1+x)dx \qquad\text{Riemann sum approximation}\\ &\int_1^2 \ln(x)dx\\ &=(x\ln(x)-x)|_1^2\\ &=(2\ln(2)-2)-(-1)\\ &=\ln(4)-1\\ &=\ln(4/e)\\ \end{array} $

Answer 4

This is a solution that a first year undergrad would be expected to give (no knowledge of Riemann integral or Stirling approximation yet beautiful in my opinion.)

We want to evaluate: $$\ell :=\lim_{n\to\infty}\frac{1}{n}\Big((n+1)(n+2)\cdots(2n)\Big)^{\frac 1n}$$

Let's define a sequence. $\displaystyle a_n=\frac{(n+1)(n+2)\ldots (2n)}{n^n}\tag*{}$ then we have $\ell = \lim\sqrt[n]{a_n}$.

Lemma: If $\lim \frac{a_{n+1}}{a_n}$ exists, it is equal to $\lim\sqrt[n]{a_n}$.

Proof of this lemma. Not related though: This wouldn't come as surprise to someone who's gone through Cauchy's $n$-th root test and D' Alembert's ratio test for convergence of infinite series. $\require{cancel}$

\begin{align}\lim \frac{a_{n+1}}{a_n}&=\frac{n^n\, \cancel{(n+2)(n+3)\ldots(2n)}(2n+1)(2n+2)}{(n+1)^{n+1}\, (n+1)\cancel{(n+2)\ldots(2n)}}\\ \\&= \lim \frac{n^n(2n+1)(2n+2)}{(n+1)^{n}(n+1)^2}\\ \\&=\lim \frac{\frac{2n+1}{n+1}\cdot\frac{2n+2}{n+1}}{\left(1+\frac 1n\right)^n}\\ \\ &=\frac{4}{e}\end{align}

$\therefore\,\boxed{\ell=\frac{4}{e}}\quad\blacksquare$