evaluate $\lim_{n\rightarrow\infty}\frac{1}{n}\sum^n_{k=1} k^{\frac{1}{k}}$ evaluate $\lim_{n\rightarrow\infty}\frac{1}{n}\sum^n_{k=1} k^{\frac{1}{k}}$
I could guess that it converges to 1 but I'm having trouble showing how

As already noted, we could appeal to the Stolz-Cesaro theorem. Instead, we go back to the basics to evaluate the limit. 
First we note that $\lim_{k\to \infty}k^{1/k}=1$.  Therefore, given any $\epsilon >0$, there exists a number $N$ such that whenever $k>N$, we have
$$1-\epsilon <k^{1/k}<1+\epsilon \tag 1$$
With that value of $\epsilon$ and $N$, we write the sum of interest as
$$\frac1n \sum_{k=1}^nk^{1/k}=\frac 1n\sum_{k=1}^Nk^{1/k}+\frac 1n \sum_{N+1}^nk^{1/k} \tag 2$$
For the fixed $N$, the first sum on the right-hand of $(2)$ goes to zero as $n\to \infty$.  For the second sum, we have using $(1)$
$$\left(1-\frac{N+1}{n}\right)(1-\epsilon)<\frac 1n \sum_{N+1}^nk^{1/k}<\left(1+\frac{N+1}{n}\right)(1+\epsilon) \tag 3$$
From $(3)$ it is clear that we can choose $n$ so large that the sum $\frac 1n \sum_{N+1}^nk^{1/k}$ is arbitrarily close to $1$.  And we are done!
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(\frac1n \sum_{k=1}^nk^{1/k}\right)=1}$$
