Methods to find $\lim\limits_{n\to\infty}\frac1n\sum\limits_{k=1}^nn^{1/k} $

AM $\ge$ GM seems sufficient to give a simple and elementary approach.

For $k \ge 2$, we have, taking $k-2$ copies of $1$ and two copies of $\sqrt{n}$, that

$$\frac{1 + 1 + \dots + 1 + \sqrt{n} + \sqrt{n}}{k} \ge n^{1/k} \ge 1$$

i.e.

$$ 1 - \frac{2}{k} + \frac{2 \sqrt{n}}{k} \ge n^{1/k} \ge 1$$

Thus

$$ (n-1) + 2(H_n - 1)(\sqrt{n} - 1) \ge \sum_{k=2}^{n} n^{1/k} \ge n-1$$

And so

$$ 2n-1 + 2(H_n - 1)(\sqrt{n} - 1) \ge \sum_{k=1}^{n} n^{1/k} \ge 2n-1$$ Where $H_n$ is the $n^{th}$ harmonic number.

Divide by $n$, and by the squeeze theorem, we have that

$$ \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n} n^{1/k} = 2 $$

This AM $\ge$ GM idea was also used in the answer here: Proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$ and a slight variant here: How to prove $\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n} - 1) = 0$? and

HINT: Looking at the sum, there are two major sources of contribution - the first few terms are large, but there are also lots of small terms on the tail that add up. So we must bound them separately, as any bounding of all terms at once will be too coarse. So separate the first $m$ terms from the rest, and estimate each part. Then look back at how to choose the $m$ so as to obtain a decent bound.

If you need more details than the hint, this page goes through the details of the method I proposed, while this takes a different approach entirely.

I made a mistake before. Now I want to fix it. I looked at the link posted above and the solution is very long and hard to follow. Please look at the solution below to see if it is correct. Note that $$ \frac{1}{n}\sum_{k=1}^nn^{1/k}=1+\frac{1}{n}\sum_{k=2}^{n}n^{1/k}. $$ Now we will show that \begin{eqnarray} \tag{1} \lim_{n\to\infty}\frac{1}{n}\sum_{k=2}^{n}n^{1/k}=1. \end{eqnarray} Let $a_n=\sum_{k=2}^{n}n^{1/k}$ and $b_n=n$. By the Stolz-Cesaro theorem, \begin{eqnarray*} \lim_{n\to\infty}\frac{1}{n}\sum_{k=2}^{n}n^{1/k}&=&\lim_{n\to\infty}\frac{a_n}{b_n}\\ &=&\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\\ &=&\lim_{n\to\infty}(a_{n+1}-a_n)\\ &=&\lim_{n\to\infty}\sum_{k=2}^{n+1}(n+1)^{1/k}-\sum_{k=2}^{n}n^{1/k}\\ &=&\lim_{n\to\infty}\left(\sum_{k=2}^{n}\left[(n+1)^{1/k}-n^{1/k}\right]+(n+1)^{1/(n+1)}\right)\\ &=&\lim_{n\to\infty}\sum_{k=2}^{n}\left[(n+1)^{1/k}-n^{1/k}\right]+1. \end{eqnarray*} Note that by the Mean Value theorem, it is easy to check $$ (n+1)^{1/k}-n^{1/k}\le \frac{1}{kn^\frac{k-1}{k}}\le\frac{1}{kn^{1/2}} $$ and hence \begin{eqnarray*} 0<\sum_{k=2}^{n}\left[(n+1)^{1/k}-n^{1/k}\right]\le\sum_{k=2}^{n}\frac{1}{kn^{1/2}}=\frac{1}{n^{1/2}}\sum_{k=2}^n\frac{1}{k}=\frac{1}{n^{1/2}}(\ln n-1+\gamma+o(1))\to 0 \end{eqnarray*} as $n\to\infty$. So (1) is true. Thus $$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nn^{1/k}=2. $$ Thank you, Aryabhata, for the suggestion to simplify the solution