2

How can I solve this limit:

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

I've tried to do it by Sandwich, but I only obtained this:

$\frac{1}{n}\sqrt[n] {(n+1)^n} \leq \frac{1}{n}\sqrt[n] {(n+1)(n+2)...(n+n)}\leq \frac{1}{n}\sqrt[n] {(n+n)^n}$

But in this way I only know that limit value is between 1 and 2.

Martin Sleziak
  • 53,687
Marcos Amorós Ibáñez
  • 77

1 Answers1

2

Let $a_n=\frac{(n+1)(n+2)...(n+n)}{n^n}$, by Stolz Theorem, we get $$\lim\limits_{n\to\infty}\sqrt[n]{a_n}=\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim\limits_{n\to\infty}\frac{(2n+1)(2n+2)}{(n+1)^2}\frac1{(1+1/n)^n}=\frac4e$$

ybtang21c
  • 451