Is $\sum i^{1/i}$ bounded?

Question

I'm trying to find the limit $$ \lim_{n\to\infty}\sum_{i=1}^n \frac{i^{1/i}}n\,. $$ I was going to say that $\lim_{n\to\infty} \frac1n=0$ and $\sum i^{1/i}$ is bounded but I can't prove it.

Answer 1

First,

$$\lim_{k\to\infty}k^{1/k}=\exp\left(\lim_{k\to\infty}\frac{\log k}k\right)=\exp0=1$$

Thus, there exists some $k_0$ such that $k> k_0$ implies $k^{1/k}\geq 1/2$. Let $$A=\sum_{k=1}^{k_0}k^{1/k}-\frac{k_0}2$$

Then, for $n> k_0$,

$$\sum_{k=1}^n k^{1/k}=\sum_{k=1}^{k_0} k^{1/k}+\sum_{k=k_0+1}^n k^{1/k}\geq A+\frac n2$$

So you can't prove that $\sum k^{1/k}$ is bounded. Simply because it is false.

Sorry for the change of notation (your $i$ is my $k$) but I only use $i$ for $\sqrt{-1}$.

Answer 2

You should use the following fact:

If $a_n\to a$, then $\,\,\dfrac{a_1+\cdots+a_n}{n}\to a,\,\,\,$ as well.

This is a consequence of Cesàro-Stolz Theorem.

In our case $$ a_n=n^{1/n}\to 1, $$ and hence $$ \frac{1^{1/1}+2^{1/2}+\cdots+n^{1/n}}{n}\to 1. $$