Prove that $a_n=\sqrt[n]{f\left(\frac{1}{n} \right)^n+f\left(\frac{2}{n} \right)^n+\dots+f\left(\frac{n}{n} \right)^n}$ is convergent 
$f$ takes positive values and is uniformly continuous. Prove that $$a_n=\sqrt[n]{f\left(\frac{1}{n} \right)^n+f\left(\frac{2}{n} \right)^n+\dots+f\left(\frac{n}{n} \right)^n}$$
  is convergent

By uniform continuity, for large enough $n$ we have $|f\left(\frac{k}{n}\right)-f\left(\frac{k+1}{n} \right)|<\epsilon$. I tried to use this in order to bound the sum, but that power $n$ makes it too large...

Let $M=\max f|_{[0,1]}$. Then for every $\epsilon>0$, we find an interval of positive length $r$ such that $f(x)>M-\frac\epsilon2$ in that interval. Note that at least $nr-1$ of the points fall into that interval.
We conclude
$$ nM^n\ge f(\tfrac1n)^n+\ldots +f(\tfrac nn)^n\ge (nr-1)(M-\tfrac\epsilon2)^n$$
and so
$$ \sqrt[n] n\cdot M\ge a_n\ge\sqrt[n]{nr-1}\cdot (M-\tfrac \epsilon2).$$
As $\sqrt[n] n\to 1$ and $\sqrt[n]{nr-1}\to 1$, we conclude that, say,  $|a_n-M|<\epsilon$ for almost all $n$, i.e., $a_n\to M$.
