Mathematical Reflections 2014, Issue 5 - Problem U314

Problem:
Prove that for any positive integer $k$,
$$\lim_{n \to \infty} \left(\dfrac{1+\sqrt[n]{2}+\ldots+\sqrt[n]{k}}{k}\right)>\dfrac{k}{e},$$
where $e$ is Euler constant.

Proposed by Ivan Borsenco.

 

Solution:
Let $f(x)=\sqrt[n]{x}$ and let $\mathcal{P}=\{[0,1],\ldots,[k-1,k]\}$ be a partition of the interval $I=[0,k]$. Since $f$ is strictly increasing on $I$, we have $$\dfrac{1}{k}\sum_{i=1}^k \sqrt[n]{i}> \dfrac{1}{k}\int_{0}^k \sqrt[n]{x} \ dx=\dfrac{n}{n+1}\sqrt[n]{k}.$$ Therefore,
$$\left(\dfrac{1}{k}\sum_{i=1}^k \sqrt[n]{i}\right)^n>k\left(1-\dfrac{1}{n+1}\right)^n$$ and taking the limit of both sides, we obtain
$$\lim_{n \to \infty}\left(\dfrac{1}{k}\sum_{i=1}^k \sqrt[n]{i}\right)^n>\dfrac{k}{e}.$$