TITLE: Methods to find $\lim\limits_{n\to\infty}\frac1n\sum\limits_{k=1}^nn^{1/k} $
QUESTION [35 upvotes]: What would you suggest here?
$$\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} n^{1/k} $$

REPLY [3 votes]: 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.