Finding the limit of $ \frac{1}{n}\sum_{k=1}^{n}\sqrt[k]{k} $ I need some help regarding a certain section in a homework question. I need to find the limit of:
$ \lim_{n\rightarrow\infty}\left(\frac{1}{n}\sum_{k=1}^{n}\sqrt[k]{k}\right)$
Now my intuition is that it converges to 1, but the whole question is centered around the number $ e $, so I'm guessing there's some connection to $ e $ and thats where I'm stuck.
Anyone have any hints?
Thanks
 A: For $k\geq N$, then $1<\sqrt[k]{k}<1+\epsilon$. So for $n\geq N$, $$\frac{1}{n}\sum_{k=N}^n1<\frac{1}{n}\sum_{k=N}^n\sqrt[k]{k}<\frac{1}{n}\sum_{k=N}^n(1+\epsilon)$$
which implies
$$\frac{1}{n}\sum_{k=1}^{N-1}\sqrt[k]{k}+\frac{1}{n}\sum_{k=N}^n1<\frac{1}{n}\sum_{k=1}^n\sqrt[k]{k}<\frac{1}{n}\sum_{k=1}^{N-1}\sqrt[k]{k}+\frac{1}{n}\sum_{k=N}^n(1+\epsilon)$$
$$\frac{A}{n}+\frac{n-N+1}{n}<\frac{1}{n}\sum_{k=1}^n\sqrt[k]{k}<\frac{A}{n}+\frac{n-N+1}{n}(1+\epsilon)$$
Now you can squeeze.
A: Hint: If $a_n\to a$ then $\frac1n\sum_{k=1}^n a_n\to a$ (Cesáro summation)
Edit:
If Cesáro cannot be used immediately, here's a quick proof:
As $a_n$ converges, it is bounded, say $|a_n|<M$ for all $n$.
Let $\epsilon>0$ be given. There exists $N$ with $|a_n-a|<\frac\epsilon2$ for all $n>N$. Then $$\left|\frac1n\sum_{k=1}^n a_n -a\right|=\left|\frac1n\sum_{k=1}^n (a_n -a)\right|\le \frac1n\sum_{k=1}^N(M+a)+\frac1n\sum_{k=N+1}^n\frac\epsilon2\le \frac Nn(M+a)+\frac\epsilon2$$
for such $n$.
Now let $N'=\max\left\{N,\left\lceil\frac{2N(M+a)}{\epsilon}\right\rceil\right\}$. Then 
$$\left|\frac1n\sum_{k=1}^n a_n -a\right|<\epsilon$$
for all $n>N'$.
A: With Stolz-Cesaro:
$$L=\lim_{n\to \infty}\left ( \frac{1}{n}\sum_{k=1}^{n}\sqrt[n]{n} \right )=\lim_{n\to \infty}\frac{\left ( \sum_{k=1}^{n}\sqrt[k]{k} \right )-\left ( \sum_{k=1}^{n-1}\sqrt[k]{k} \right )}{n-(n-1)}=\lim_{n\to \infty}\sqrt[n]{n}$$
Hence
$$\lim_{n\to\infty} \frac{\ln n}{n}=\lim_{n\to \infty}\frac{\ln(n+1)-\ln n}{(n+1)-n}=\lim_{n\to \infty}\ln\left ( 1+\frac{1}{n} \right )=0$$
$$\to\lim_{n\to \infty}\sqrt[n]{n}=1$$
Hence: $L=1$
