Límite de: $\lim\limits_{x\to \infty}\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx}$

Question

Calcular el siguiente límite ($n\in \mathbb{N}$) $$\lim_{x\to \infty}\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx}$$

Mi idea era usar la desigualdad:

$$\left(\frac{1}{n}\sum_{k=1}^{n} 1^{1/x}\right)^{nx}<\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx}<\left(\frac{1}{n}\sum_{k=1}^{n} n^{1/x}\right)^{nx} \\ \implies1<L<n^n$$

Esto le da a que el límite requerido $L$ se encuentra entre $1$$n^n$. Pero ¿cómo podemos saber su valor?

Answer 1

Desde $e^t=1+t+O\!\left(t^2\right)$, $t=\frac{\log(k)}x$ $$ \begin{align} \lim_{x\to\infty}\left(\frac1n\sum_{k=1}^nk^{1/x}\right)^{nx} &=\lim_{x\to\infty}\left(1+\frac1n\sum_{k=1}^n\frac{\log(k)}x+O\!\left(\frac1{x^2}\right)\right)^{nx}\\ &=\lim_{x\to\infty}\left(1+\frac1n\sum_{k=1}^n\frac{\log(k)}x\right)^{nx}\lim_{x\to\infty}\left(1+O\!\left(\frac1{x^2}\right)\right)^{nx}\\[3pt] &=\lim_{x\to\infty}\left(1+\frac1n\frac{\log(n!)}x\right)^{nx}\cdot1\\[9pt] &=e^{\log(n!)}\\[15pt] &=n! \end{align} $$

Answer 2

Deje $$f(h)= \ln\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h}\right)\implies f'(h) =\frac{\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h}\ln k \right)}{\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h} \right)}$$ $$f(0)= 0 ~~~and~~~~f'(0)=\left(\frac{1}{n}\sum\limits_{k=1}^{n} \ln k\right)=\color{blue}{ \frac{1}{n}\ln \left(n!\right) } $$

Deje $x=1/h$, entonces tenemos

$$\lim_{x\to \infty}\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx} =\lim_{h\to 0}\exp\left(\frac{n}{h}\ln\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h}\right)\right)\\ =\lim_{h\to 0}\exp\left(n\frac{f(h)}{h}\right) =\color{red}{\exp\left(nf'(0)\right) }=\color{red}{n! } $$