Methods to find $\lim\limits_{n\to\infty}\frac1n\sum\limits_{k=1}^nn^{1/k} $ What would you suggest here?
$$\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} n^{1/k} $$
 A: 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
A: Using Bernoulli's Inequality, we get for $k\ge\log(n)$
$$
\frac1n\ge\left(1-\frac{\log(n)}{k}\right)^k\tag{1}
$$
from which it is simple to derive
$$
n^{1/k}-1\le\frac{\log(n)}{k-\log(n)}\tag{2}
$$
Just multiplying the number of terms by the largest term, we get
$$
\begin{align}
\sum_{k=2}^{\lfloor\log(n)\rfloor+1}\left(n^{1/k}-1\right)
&\le(\sqrt{n}-1)\lfloor\log(n)\rfloor\\
&\le\sqrt{n}\,\log(n)\tag{3}
\end{align}
$$
Using $(2)$, we get
$$
\begin{align}
\sum_{k=\lfloor\log(n)\rfloor+2}^n\left(n^{1/k}-1\right)
&\le\sum_{k=\lfloor\log(n)\rfloor+2}^n\frac{\log(n)}{k-\log(n)}\\
&\le\log(n)(\log(n)+\gamma)\\[8pt]
&\le(\log(n)+\gamma)^2\tag{4}
\end{align}
$$
Combine $(3)$ and $(4)$ to get
$$
\lim_{n\to\infty}\frac1n\sum_{k=2}^n\left(n^{1/k}-1\right)=0\tag{5}
$$
Therefore,
$$
\lim_{n\to\infty}\frac1n\sum_{k=1}^nn^{1/k}=2\tag{6}
$$
A: 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
A: I tried to post this answer to the question How to solve $\lim_{n \to \infty} \frac{n+n^{\frac{1}{2}}+n^{\frac{1}{3}}+...+n^{\frac{1}{n}}}{n}$. Then I noticed that it was a duplicate of the present one. Having seen that my solution method did not yet appear here I add it.
Answer
The limit in question is 2.
The graph depicts the partial sums for $n= 1..500$. 

Notice that there is a maxiumum at $n=13$ and that the limit $s(\infty)=2$ is approached from above and very slowly.
Derivation
We have to calculate the limit of this sum
$$s_{n} = \sum_{k=1}^n c_{k}\tag{1a}$$
Where
$$c_{k} = c(k) = n^{\frac{1}{k}-1}\tag{1b}$$
We use partial summation in the integral form (see my answer to Does $\sum_{n=1}^\infty \frac{\cos{(\sqrt{n})}}{n}$ converge?) which gives us
$$s_{n} = i_0 + i_1(n) - i_2(n)\tag{2a}$$
Where
$$i_0 = c_1 = 1\tag{2b}$$
$$i_1(n) = \int_{1}^n c(x)\,dx\tag{2c}$$
$$i_2(n) = - \int_{1}^n \{x\} c'(x)\,dx\tag{2d}$$
where $\{x\}$ is the fractional part of $x$.
The first integral can be done (Mathematica) leading to
$$i_1(n) = -\frac{\log (n) \text{Ei}\left(\frac{\log (n)}{n}\right)}{n}+\frac{\log (n) \text{Ei}(\log (n))}{n}+n^{1/n}-1\tag{3a}$$
Where $\text{Ei}$ is the exponential integral function.
In the limit we find 
$$i_1(\infty) = \lim_{n\to \infty } \, i_1(n) = 1\tag{3b}$$
The second integral will be shown to vanish in the limit, so that we find the announced answer
$$s_{\infty}  = i_0 + i_1(\infty) = 2\tag{4}$$
Indeed, since 
$$c'(x) = -\left(\frac{\log (n)}{n}\right)\left(\frac{n^{\frac{1}{x}}}{x^2}\right)\tag{1c}$$ 
the second integral, after transforming $x\to\frac{1}{y}$, becomes
$$i_2(n) = \frac{\log (n)}{n}\int_{\frac{1}{n}}^1 \{\frac{1}{y}\} n^{y} \,dy \tag{5}$$
Observing that 
$$0 \lt \{\frac{1}{y}\} \le \left(\frac{1}{y}-1 \right)\text{ for} \;\;0 \lt y \lt 1 \tag{6}$$ 
we have
$$i_{2}(n) \le \frac{\log (n)}{n}\int_{\frac{1}{n}}^1 \left(\frac{1}{y} -1\right) n^{y} \,dy \\
= \left(\frac{\log (n)}{n}\right) \left(\text{Ei}\left(\frac{\log (n)}{n}\right)+\text{Ei}(\log (n))+\frac{n^{1/n}-n}{\log (n)}\right)\tag{7}$$
The asymptotic behaviour of this expression is
$$\frac{1}{\log (n)}+O\left(\frac{\log ^2(n)}{n}\right)\tag{8}$$
This goes to zero, albeit very slowly.
Hence we have obtained the inequality
$$0 \lt i_2(n) \lt \frac{1}{\log (n)}+O\left(\frac{\log ^2(n)}{n}\right)$$
Which means that in the limit $i_{2}(\infty)=0$.
EDIT 19.10.18
Simplification of limit of the integral $i_1$.
We have 
$$i_1(n) = \int_{1}^n c(x)\,dx= \frac{1}{n}\int_{1}^n n^{\frac{1}{x}}\,dx= \frac{1}{n}\int_{\frac{1}{n}}^1 n^{y}\frac{1}{y^2}\,dy$$
Since the main contribution comes from $y\simeq 0$ we can expand $n^y$ into a power series and write
$$i_1(n) = \frac{1}{n}\int_{\frac{1}{n}}^1 \left(1+y \log(n) + \frac{1}{2}y^2 \log(n)^2+...\right)\frac{1}{y^2}\,dy\\
=\frac{1}{n}\left(\left( n-1\right)-\left(\log \left(\frac{1}{n}\right) \log (n)\right)+ \left(\frac{1}{2} \left(1-\frac{1}{n}\right) \log ^2(n)\right)+... \right)$$
Hence we find for the limit 
$$i_1(\infty) = \lim_{n\to \infty } \,i_1(n)=1$$
Unfortunately, things are not as easy for $i_2$, so that we have to stick to the formula (7) etc.
A: 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. 
