Asymptotics of sum without Euler-Maclaurin I want to show that
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n n^{1/k} = 2$$
I found the limit with Euler-Maclaurin but this was messy involving integral $\int_1^nn^{1/x}$dx.
What is an easier way using squeeze or other technique?
 A: $n^{1/k}=\exp\frac{\log n}{k}=1+\frac{\log n}{k}+\sum_{m\geq 2}\frac{1}{m!}\cdot\frac{\left(\log n\right)^m}{k^m}$. It follows that:
$$ \sum_{k=1}^{n}n^{1/k} = n+ H_n \log(n) + \sum_{m\geq 2}\frac{H_n^{(m)}}{m!}\left(\log n\right)^m. $$
Since $H_n^{(m)}\geq 1$ and $H_n\geq \log(n)$, we trivially have
$$ \sum_{k=1}^{n}n^{1/k}\geq n +\log^2(n)+\sum_{m\geq 2}\frac{(\log n)^m}{m!} = 2n+O(\log^2(n)).$$
On the other hand $H_n^{(m)}\leq \zeta(m)\leq 1+\int_{1}^{+\infty}\frac{dx}{x^m}=1+\frac{1}{m-1}$, from which it is simple to show that $\sum_{k=1}^{n}n^{1/k}\leq 2n+O(\log^3(n))$. The claim then follows by squeezing.
A: Let $S_n=\sum_{k=1}^n n^{1/k}$. 
Clearly, 
$$S_n \ge  n+  \sum_{k=2}^n 1 =n+(n-1) = 2n-1.$$ 
On the other hand, 
$$S_n =n+ \underset{(I)}{\underbrace{\sum_{k=2}^{k_1} n^{1/k} }}+\underset{(II)}{\underbrace{\sum_{k=k_1+1}^n n^{1/k}}}.$$ 
Choose $k_1 = \lfloor n^{1/4}\rfloor $ (and assume that $n$ is large enough so that $n^{1/4}\ge 2$).  Then 
\begin{align*} 
(I)&\le n^{1/2}\times n^{1/4}\mbox{, and}\\
(II)&\le n^{1/ n^{1/4}}\times n.
\end{align*}
Observe that $n^{1/n^{1/4}} = ((n^{1/4})^{1/n^{1/4}})^4\to 1$. Putting it all together gives 
$$S_n \le n + n^{3/4} + (1+o(1))n=(2+o(1))n.$$  
