Limit of $\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$

Question

Compute the limit $$ \lim_{n \to \infty} \frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n $$

How can this be done? The best I could do was rewrite the limit as $$ \lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} $$

Following that log suggestion in the comments below: \begin{align} &\ln \left(\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \ln \left( \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \left(\ln \left(\frac{n+1}n \right)^{\frac 1n} + \ln\left(\frac{n+2}n \right)^{\frac 1n} + \cdots + \ln\left(\frac{n+n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \sum_{i=1}^n \ln \left(\frac{n+i}n \right)^{\frac 1n} \\ &= \lim_{n \to \infty} \frac 1n \sum_{i=1}^n \ln \left(1 + \frac in \right) \\ &= \int_1^2 \ln x \, dx \\ &= (x \ln x - x)\vert_1^2 \\ &= (2 \ln 2 - 2)-(1 \ln 1-1) \\ &= (\ln 4-2)-(0-1) \\ &= \ln 4-1 \\ &= \ln 4 - \ln e \\ &= \ln \left( \frac 4e \right) \end{align}

but I read from somewhere that the answer should be $\frac 4e$.

Answer 1

If you know Sterling's Approximation: $$ n! \sim \left(\frac{n}{e}\right)^n \sqrt{2\pi n}, $$ then you could approach it as follows:

$$ \lim_{n \to \infty} \frac{\sqrt[n]{(n + 1)(n + 2)\cdots (2n)}}{n} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{(2n)!}{n!}} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{\left(\frac{2n}{e}\right)^{2n} \sqrt{4\pi n}}{\left(\frac{n}{e}\right)^n \sqrt{2\pi n}}} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{2^{2n} \left(\frac{n}{e}\right)^n \sqrt{2}} = \lim_{n \to \infty} \frac{1}{n} \cdot 2^2 \cdot \frac{n}{e} \cdot 2^{\frac{1}{2n}} = \frac{4}{e}. $$

Answer 2

Hint: let $a_n=(n+1)\cdots(2n)/n^n$, and use the fact that $a_{n+1}/a_n\to L$ implies $\sqrt[n]{a_n}\to L$ (i.e., the ratio test implies the root test).

Answer 3

Now take a logarithm and realize you have just gotten a Riemann sum. (Interval: $[1,2]$, widths: $1/n$, heights $\ln(1+k/n)$.)

Answer 4

Note that $\displaystyle\frac{\sqrt[n]{n+1}}{n}\neq\left(\frac{n+1}{n}\right)^{1/n}$

Also, this is just for clarification since some nice suggestions have already been given but also because the question mark remains over your last equality

With this is mind and going from your first step,

$$S=\lim_{n\to\infty}\frac{\sqrt[n]{(n+1)(n+2)...(2n)}}{n}=\lim_{n\to\infty}\left(\frac{1}{n}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)\right)$$

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

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

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\operatorname{ln}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)-\operatorname{ln}(n)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\operatorname{ln}\left((n+1)^{1/n}\right)+\operatorname{ln}\left((n+2)^{1/n}\right)+...+\operatorname{ln}\left((2n)^{1/n}\right)-\operatorname{ln}(n)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\sum_{p=1}^n\left(\operatorname{ln}\left((n+p)^{1/n}\right)\right)-\operatorname{ln}(n)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\sum_{p=1}^n\left(\frac{1}{n}\operatorname{ln}(n+p)\right)-\operatorname{ln}(n)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\left(\operatorname{ln}(n+p)\right)-\operatorname{ln}(n)\right)$$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\left(\operatorname{ln}\left(1+\frac{p}{n}\right)+\operatorname{ln}(n)\right)-\operatorname{ln}(n)\right)$$

This $+\operatorname{ln}(n)$ is independent of $p$ and summed $n$ times, then divided by $n$, so $\operatorname{ln}(n)-\operatorname{ln}(n)=0$

$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\operatorname{ln}\left(1+\frac{p}{n}\right)\right)$$

This is what Eric meant and the argument varies from $1$ to $2$

It becomes a little clearer if we let $m = 1/n$

$$\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\operatorname{ln}\left(1+\frac{p}{n}\right)\right)=\lim_{m\to\ 0}\left(\left(\sum_{p=1}^{1/m}\operatorname{ln}(1+pm)\right)m\right)=\int_1^2\operatorname{ln}(x)dx=(x\operatorname{ln}(x)-x)\Big\rvert_1^2$$

$$\therefore \operatorname{ln}(S)=2(\operatorname{ln}(2)-1)+1=\operatorname{ln}(4)-1=\operatorname{ln}\left(\frac{4}{e}\right)$$

Finally

$$S=\frac{4}{e}$$

Answer 5

If you accept using the following limit that has been asked and proved (for example here) many times here on MSE

• $\lim_{n\rightarrow \infty}\frac{\sqrt[n]{n!}}{n} = \frac{1}{e}$,

then your limit is easily derived:

$$\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n = \frac{ (2n!)^{\frac{1}{n}} }{ (n!)^{\frac{1}{n}}\cdot n} = 4\frac{n}{ (n!)^{\frac{1}{n}}} \left( \frac{ (2n!)^{\frac{1}{2n}} }{ 2n} \right)^2 \stackrel{n\rightarrow\infty}{\longrightarrow}4\cdot e \cdot \frac{1}{e^2}=\frac{4}{e}$$