Asymptotics of sum without Euler-Maclaurin

Question

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?

Answer 1

$n^{1/k}=\exp\frac{\log n}{k}=1+\frac{\log n}{k}+\sum_{m\geq 2}\frac{1}{m!}\cdot\frac{\left(\log n\right)^m}{k^m}$. 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.

Answer 2

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

$$S_n \le n + n^{3/4} + (1+o(1))n=(2+o(1))n.$$