How to find a limit of this sequence: $\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \frac{1}{\sqrt{kn}}$

Question

How to fing a limit $$\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{\sqrt{kn}}?$$ I had only two minds about this. First of them was that it looks like $\frac{1^k +…+n^k}{n^{k+1}}$ which limit is $\frac{1}{k+1}$. The second is to look at square of $1+\frac{1}{\sqrt{2}} + .. + \frac{1}{\sqrt{n}}$, but it is a kind of a monster. I think, I need a formula for $1+\frac{1}{\sqrt{2}} + .. + \frac{1}{\sqrt{n}}$, but I don’t know it.

Answer 1

HINT

We have that

$$\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{\sqrt{kn}}=\lim_{n \to \infty} \frac1n\sum_{k=1}^n \frac{1}{\sqrt{\frac k n}}$$

then refer to

• Perfect understanding of Riemann Sums

Answer 2

If you manage to prove that $$ H_n^{(1/2)}=\sum_{k=1}^{n}\frac{1}{\sqrt{k}} = 2\sqrt{n}(1+o(1)) \tag{1}$$ you are done, the limit is $2$. On the other hand

$$ \frac{H_{n+1}^{(1/2)}-H_n^{(1/2)}}{2\sqrt{n+1}-2\sqrt{n}}=\frac{2\sqrt{n+1}+2\sqrt{n}}{4\sqrt{n+1}}\to 1\tag{2}$$ hence $(1)$ is a straightforward consequence of Cesàro-Stolz.
Through creative telescoping or just induction you may actually prove the stronger bound $$2\sqrt{n}-\tfrac{3}{2}\leq H_n^{(1/2)}\leq 2\sqrt{n}-1.\tag{3}$$