How to calculate this limit? $$\lim_{n\to \infty}\frac{\sum_{k=1}^nn^{\frac{1}{k}}}{n}=\lim_{n\to \infty}\frac{n+n^{\frac{1}{2}}+n^{\frac{1}{3}}+\cdots+n^{\frac{1}{n}}}{n}$$ By computer it seems the limit is $2$. The lower bound is easy. But I don't know how to calculate the upper bound. I have tried Stolz theorem, then we need to estimate $$c_n=1+(n+1)^{\frac{1}{n+1}}+\sum_{k=2}^n((n+1)^{\frac{1}{k}}-n^{\frac{1}{k}})$$ since $k$ is associate to $n$, I failed to estimate $\sum_{k=2}^n((n+1)^{\frac{1}{k}}-n^{\frac{1}{k}})$. Besides, this limit looks like the Riemann sum but I didn't find the form and function. So cound you please give some tips?
Or if there is the same question please share the link, thanks.
