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.
 A: mrf has the main idea.  But since the integral is improper (as mrf notes) some care is required.  Here is one way, using the Lebesgue theory...
Let $f(x) = 1/\sqrt{x}$.  For fixed $n$, let $g_n$ be defined by $g_n(x) = \sqrt{n/k}$ for $(k-1)/n < x \le k/n$.  Then $0 < g_n(x) \le f(x)$ and $g_n(x) \to f(x)$ on $(0,1]$.  Since $f$ is Lebesgue integrable on $(0,1]$, we have by the dominated convergence theorem $\int_0^1 g_n \to \int_0^1 f$.  That is:
$$
\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{n}{k}} \to \int_0^1 \frac{dx}{\sqrt{x}}
$$
ASIDE 
Monthly problem 11376 has an example of how blindly saying "Riemann sum" can lead one astray.  The solution is on p. 283 of the March, 2010 issue.  The problem defines
$$
S_n(a) = \sum_{an \lt k \le (a+1)n}\frac{1}{\sqrt{kn-an^2}\;}
$$
for real $a$ and positive integer $n$, and asks for which $a$ does $\lim_{n \to \infty} S_n(a)$ exist.  Many solvers noted that $S_n(a)$ is a Riemann sum for
$$
\int_a^{a+1} \frac{dx}{\sqrt{x-a}\;} = 2
$$
and then carelessly concluded that $S_n(a) \to 2$ for all $a$.  But, in fact, as the published solution shows, $S_n(a)$ converges if and only if $a$ is rational.  
The problem here is the case $a=0$, and fortunately $0$ is rational.
A: Think about $$\int_0^1 f(x)dx,~~~f(x)=\frac{1}{\sqrt{x}}
$$
Indeed: $$\int_a^b f(x)dx=\lim_{n\to\infty}\sum_{i=1}^nf\left(a+\frac{b-a}{n}i\right)\left(\frac{b-a}{n}\right)$$
A: Your sum can be interpreted as a Riemann sum:
$$\sum_{k=1}^n \frac{1}{\sqrt{kn}} = \frac1n \sum_{k=1}^n \sqrt{\frac{n}{k}}.
$$
Let $f(x) = 1/\sqrt{x}$ and let $x_k = k/n$. Then 
$$\sum_{k=1}^n \frac{1}{\sqrt{kn}} = \frac1n \sum_{k=1}^n \sqrt{\frac{n}{k}} = \frac1n \sum_{k=1}^n f(x_k) \to \int_0^1 f(x)\,dx
$$
as $n \to \infty$.
(Since the integral is improper, a little care is needed to justify the last step.)
A: 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,
$$\left|\;\sum_{k=1}^n \frac{1}{\sqrt{kn}} - 2\;\right| \le 2 \left(1+\frac{1}{\sqrt{n}} -\sqrt{1+\frac{1}{n}}\right) < \frac{2}{\sqrt{n}}
\quad\implies\quad \lim_{n\to\infty} \sum_{k=1}^n \frac{1}{\sqrt{kn}} = 2.$$
