Evaluating Limit Question $\lim\limits_{n\to \infty}\ \frac{1+\sqrt[2]{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}=1$? Does $\displaystyle\lim_{n\to \infty}\ \frac{1+\sqrt[2]{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}=1$? 
Thanks a lot for your time and help.
 A: The limit immediately evaluates to $1$ after applying Stolz-Caesaro.
If one wants a more precise estimate, Euler-Maclaurin and $\sqrt[n]{n} = e^{(\log n)/n} $ gives: $$\sum_{k=1}^n \sqrt[k]{k} = \sum_{k=1}^n \left(1+ \frac{\log k}{k} + \mathcal{O}\left( \frac{\log^2 k}{k^2} \right) \right) = n + \frac{\log^2 n}{2} + \frac{\log n}{2n} + \mathcal{o}\left(\frac{\log n}{n}\right).$$
A: The answer is $1.$ You can use the following result:
If $\lim\limits_{n\to\infty }a_n=a$, then we have 
$$\lim_{n\to\infty} \frac{a_1+a_2+\cdots+a_n}{n}=a.$$
And you can prove this result by the definition of limits of sequences. Note that $\lim\limits_{n\to\infty}\sqrt[n]{n}=1.$
A: For a completely elementary proof, you can use the elementary theorem(IMO) that if $a_n \to a$ then $\frac{1}{n} \sum_{k=1}^{n} a_k \to a$ as mentioned in the comments and in the other answers.
We can also do using the squeeze (but I don't see a valid proof in the comments yet).
As in my answer here: How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$?
By using $\text{AM} \ge \text{GM}$ on $k-2$ ones, $\sqrt{k}$, $\sqrt{k}$ that, for $k \ge 3$,
$$ 1  + \frac{2}{\sqrt{k}} \ge 1 - \frac{2}{k} +\frac{2}{\sqrt{k}} \ge k^{1/k} \ge 1 $$
Now we have that (using the mean value theorem, or otherwise)
$$ \frac{1}{2\sqrt{k+1}} \lt \sqrt{k+1} - \sqrt{k} \lt \frac{1}{2\sqrt{k}}$$
Adding gives us that
$$ \sum_{k=3}^{n} \frac{2}{\sqrt{k}} \le 4\sqrt{n} + 2$$
Thus
$$ n + 4\sqrt{n} + 10 \ge \sum_{k=1}^n k^{1/k} \ge n$$
And so 
$$1  + \frac{4}{\sqrt{n}} + \frac{10}{n} \ge \frac{1}{n}\sum_{k=1}^n k^{1/k} \ge 1$$
and now we can apply the squeeze.
A: Set $a_n=1+\root 2\of 2+\root3\of 3+\cdots +\root n\of n$. 
Here is a proof based on the facts that
$\ \ \ \ \ $1) $\lim\limits_{n\rightarrow\infty} \root n\of n=1$ 
and 
$\ \ \ \ \ \ $2) $(\root n\of n)$ is decreasing for $n\ge3$.
Trivially, ${a_n\over n}\ge 1$ for all $n$; thus,
$$\tag{1}
\liminf_{n\rightarrow\infty}{a_n\over n}\ge 1.
$$
By 2), we have for a fixed $M\ge 3$  and $k$ a positive integer:
$$\eqalign{
{a_{M+k}\over M+k}
\ \ &\ \ ={a_M\over M+k}+{ (M+1)^{1/M+1} + (M+2)^{1/M+2}+\cdots+(M+k)^{M+k}\over M+k }\cr
\ \ &\ \ \le  {a_M\over M+k}+{ k(M+1)^{1/M+1} \over M+k}\cr
 & \buildrel{ k\rightarrow\infty}\over\longrightarrow \ \ 0+{(M+1)^{M+1}}.\phantom{M\over M}
}
$$
From this it follows that $\limsup\limits_{n\rightarrow\infty} {a_n\over n}\le (M+1)^{1/M+1}$ for all $M\ge3$; and, thus by 1), we have  $\limsup\limits_{n\rightarrow\infty} {a_n\over n}\le1$. Combining this result with $(1)$, we have $\lim\limits_{n\rightarrow\infty }{a_n\over n}=1$.
This is just a modification of the proof of the Cesaro  theorem found here.
The decreasing condition 2) used in the above is not needed, and, alternatively, you could just use the proof in the link. 
