<s>[INST] How to evaluate the sum $\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{2n}}+\cdots+\frac{1}{\sqrt{n^2}}$ when $n$ grows? I need help with the following limit $$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{\sqrt{kn}}$$
Thanks.
 [/INST] Notice for large $n$, we expect $\displaystyle \sum_{k=1}^n \frac{1}{\sqrt{k}}
\text{ behave like }\int_1^n \frac{dx}{\sqrt{x}} \sim 2\sqrt{n}$. This suggests
$$\frac{1}{\sqrt{k}} \sim \int_{k-1/2}^{k+1/2} \frac{dx}{\sqrt{x}} \sim 2\left( \sqrt{k+\frac12} - \sqrt{k-\frac12}\right)$$ and the terms $\displaystyle \frac{1}{\sqrt{kn}}$ 
in the summands is close to something "telescopable". To make this idea concrete, we observe:
$$\begin{align}
\sum_{k=1}^n\frac{1}{\sqrt{kn}} \ge &
\sum_{k=1}^n \frac{2}{\sqrt{n}(\sqrt{k+1}+\sqrt{k})} 
= \frac{2}{\sqrt{n}} \sum_{k=1}^n(\sqrt{k+1}-\sqrt{k})
= 2 \Big(\sqrt{1+\frac{1}{n}} - \frac{1}{\sqrt{n}}\Big)\\
\sum_{k=1}^n\frac{1}{\sqrt{kn}} \le &
\sum_{k=1}^n \frac{2}{\sqrt{n}(\sqrt{k}+\sqrt{k-1})} 
= \frac{2}{\sqrt{n}}\sum_{k=1}^n(\sqrt{k}-\sqrt{k-1}) = 2
\end{align}$$
As a result,
$$\left|\;\sum_{k=1}^n \frac{1}{\sqrt{kn}} - 2\;\right| \le 2 \left(1+\frac{1}{\sqrt{n}} -\sqrt{1+\frac{1}{n}}\right) < \frac{2}{\sqrt{n}}
\quad\implies\quad \lim_{n\to\infty} \sum_{k=1}^n \frac{1}{\sqrt{kn}} = 2.$$
</s>