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...
 A: Hint: Well, if we have a finite number of positive terms $a_1,a_2,\ldots,a_m$ then
$$ \lim_{n\to +\infty}\sqrt[n]{a_1^n+a_2^n+\ldots+a_m^n} = \max(a_1,a_2,\ldots,a_m)$$
is well-known and not difficult to prove. If $f$ is positive and continuous on $[0,1]$, by letting $M=\sup_{[0,1]}f(x)=\max_{[0,1]}f(x)$ we have
$$(M-\varepsilon_n)^n\leq f\left(\tfrac{1}{n}\right)^n+f\left(\tfrac{2}{n}\right)^n+\ldots+ f\left(\tfrac{n}{n}\right)^n \leq n M^n$$
and $\lim_{n\to +\infty}\sqrt[n]{n}=1$, so the situation is essentially the same.
A: 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., $a_n\to M$.
A: If $f[0,1]\to [0,\infty)$, continuous, then
$$
\lim_{n\to\infty}\sqrt[n]
{\,
f\left(\frac{1}{n}\right)^{n^{\phantom{-}}}+
f\left(\frac{2}{n}\right)^{n^{\phantom{|}}}+\cdots+
f\left(\frac{n}{n}\right)^{n^{\phantom{|}}}
}=\max_{x\in [0,1]}f(x).
$$
Proof. Let $x_0\in[0,1]$, such that $$f(x_0)=\max_{x\in [0,1]}f(x)=M$$ and $\varepsilon>0$ arbitrary. Then there is a $\delta>0$, such that if $|x-x_0|<\delta$,
then $\,|\,f(x)-f(x_0)|<\varepsilon$. Thus, if
$$
n\ge n_1=\left\lfloor\frac{1}{\delta}\right\rfloor+1\quad\Longrightarrow\quad
\frac{1}{n}<\delta,
$$
and hence, for $n\ge n_1$, the distance of one of the points $\frac{1}{n},\frac{2}{n},\ldots,\frac{n}{n}$ is less that $\delta$ and thus
among the values $f\left(\frac{1}{n}\right),f\left(\frac{2}{n}\right),\ldots,f\left(\frac{n}{n}\right)$, there exists one,  say $f\left(\frac{k}{n}\right)$, which differs less that $\varepsilon$ from $f(x_0)$, i.e.
$$
M-\varepsilon< f\left(\frac{k}{n}\right)\le M.
$$
Next, as $Mn^{1/n}\to M$, there exists an $n_2$, such that if $n\ge n_2$, then $Mn^{1/n}<M+\varepsilon$. 
Thus, for every $n\ge n_0=\max\{n_1,n_2\}$
$$
(M-\varepsilon)^n<f\left(\frac{1}{n}\right)^{n^{\phantom{-}}}+
f\left(\frac{2}{n}\right)^{n^{\phantom{|}}}+\cdots+
f\left(\frac{n}{n}\right)^{n^{\phantom{|}}}\le nM^n
$$
and thus
$$
M-\varepsilon<\sqrt[n]{f\left(\frac{1}{n}\right)^{n^{\phantom{-}}}+
f\left(\frac{2}{n}\right)^{n^{\phantom{|}}}+\cdots+
f\left(\frac{n}{n}\right)^{n^{\phantom{|}}}}\le Mn^{1/n}<M+\varepsilon
$$
Altogether, for every $\varepsilon>0$, there exists an $n_0\in\mathbb N$, such that
$$
n\ge n_0\quad\Longrightarrow\quad \left|\sqrt[n]{f\left(\frac{1}{n}\right)^{n^{\phantom{-}}}+
f\left(\frac{2}{n}\right)^{n^{\phantom{|}}}+\cdots+
f\left(\frac{n}{n}\right)^{n^{\phantom{|}}}}-M\right|<\varepsilon.
$$
QED
