2

Evaluate the limit: $\lim_{n\to \infty} \frac{\left[(n+1)(n+2)\ldots(n+n)\right]^{1/n}}{n}$

I can write $$\lim_{n\to \infty} \frac{[(n+1)(n+2)\ldots(n+n)]^{1/n}}{n}$$ $$=\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{1/n}\left(1+\frac{2}{n}\right)^{1/n}\ldots \left(1+\frac{n}{n}\right)^{1/n}$$$$=e\times \sqrt{e}\times \sqrt[3]{e}\times \ldots \times \sqrt[n-1]{e}\times 1$$.

I don't know how to evaluate the final product.

Arctic Char
  • 16,007
Miz
  • 2,739
  • 6
    @Miz What you do doesn't make much sense, how do you get from the first to the second line? How do you get from the second to the third? You can't employ the product rule for limits here as the length of the product itself depends on $n$. Moreover, $$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{1/n} \neq e$$ – Richard Feb 03 '23 at 11:28

3 Answers3

7

Taking the logarithm gives $$ \frac{1}{n}\sum_{k=1}^n\log(n+k)-\log n=\frac{1}{n}\sum_{k=1}^n\log\left(1+\frac{k}{n}\right). $$ This is a Riemann sum which converges to $\int_0^1 \log(1+x)dx=\log 4 -1$. The desired limit is therefore $\frac{4}{e}$.

Tuvasbien
  • 8,907
3

We have, using Stirling's formula: $$\begin{split}\frac{\prod_{k=1}^n (n+k)}{n^n} = \frac{(2n)!}{n^n \cdot n!} &\underset{n \to \infty}{\sim} \frac{ \left(\frac{2n}{e}\right)^{2n} \sqrt{2\pi 2n}}{n^n \cdot \left(\frac{n}{e}\right)^{n} \sqrt{2\pi n}}\\ &\underset{n \to \infty}{\sim} \frac{4^n \sqrt{2}}{e^n}\end{split}$$ As such, since putting both sides to the power $\frac{1}{n}$ is allowed, we have that our given limit is $\frac{4}{e}$.

That is allowed because if $u_n = e_n v_n$ with $e_n \to 1$, then $u_n^{1/n} = e_n^{1/n} v_n^{1/n}$.
But for all big enough $n$, $\frac{1}{2} \leq e_n \leq \frac{3}{2}$, so for those same $n$: $\left(\frac{1}{2}\right)^{1/n} \leq e_n^{1/n} \leq \left(\frac{3}{2}\right)^{1/n}$, and both expressions on the sides tend to $1$, so the middle one too by the squeeze theorem.

Bruno B
  • 5,027
3

Using Pochhammer symbol $$\prod_{i=1}^n (n+i)=(n+1)_n=\frac{2^{2 n} }{\sqrt{\pi }}\Gamma \left(n+\frac{1}{2}\right)$$ So, you want to compute $$A_n=\frac 1n\left(\frac{2^{2 n} }{\sqrt{\pi }}\Gamma \left(n+\frac{1}{2}\right)\right)^{\frac 1n}$$

Taking logarithms and using Stirling expansion $$\log(A_n)=(2 \log (2)-1)+\frac{\log (2)}{2 n}-\frac{1}{24 n^2}+O\left(\frac{1}{n^4}\right)$$ $$A_n=e^{\log(A_n)}=\frac 4e \left(1+\frac{\log (2)}{2 n}+\frac{3 \log ^2(2)-1}{24 n^2}+O\left(\frac{1}{n^3}\right)\right)$$ whose relative error is less than $0.1$% as soon as $n \geq 2$.

Claude Leibovici
  • 260,315