Find the value of $\lim_{n\to\infty}\frac{1+2^{1/2}+3^{1/3}+\cdots+n^{1/n}}{n}$

Question

The given problem is a problem of calculus. I have tried this problem only to find that the numerator is a divergent series. No idea how to do this problem

$$\lim_{n\to\infty}\frac{1+2^{1/2}+3^{1/3}+\cdots+n^{1/n}}{n}$$

Answer 1

HINT

By Stolz-Cesaro

$$\lim_{n \to \infty} \frac{\sum_{k=1}^n k^\frac1k}n=\lim_{n\to\infty}\quad (n+1)^{\frac1{n+1}}=1$$

Answer 2

You can prove in general that if $a_n\to a$ as $n\to \infty$, then $$ \sum_{i = 1}^n\frac{a_i}{n}\to a $$ as well. Now note that $\sqrt[n]{n}\to 1$, and you're done.