Is this sequence convergent? If so, then how to find the limit? For each natural number $n$, let
$$ x_n \colon= \frac{1}{\sqrt{n}} \left( \frac{1}{\sqrt{1}}  + \frac{1}{\sqrt{2}}+ \cdots + \frac{1}{\sqrt{n}} \right). $$
Then is the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ convergent? And if so, then how to find the limit of this sequence? 
I'm just not sure how to proceed. 
 A: METHODOLOGY $1$:  USE CREATIVE TELESCOPING 
since $\frac12(\sqrt {k-1}+\sqrt k)\le \sqrt k\le \frac12(\sqrt {k+1}+\sqrt k)$, we can use creative telescoping to write
$$\sum_{k=1}^n2\left(\sqrt{k+1}-\sqrt k\right)\le\sum_{k=1}^n\frac{1}{\sqrt k}\le 1+\sum_{k=2}^n 2\left(\sqrt k-\sqrt{k-1}\right)$$
Evaluating the telescoping terms reveals
$$2(\sqrt {n+1}-1)\le \sum_{k=1}^n\frac{1}{\sqrt k}\le 2\sqrt n-1$$
whence dividing by $\sqrt n$ and applying the squeeze theorem yields the result
$$\lim_{n\to \infty }\frac1{\sqrt  n}\sum_{k=1}^n \frac1{\sqrt k}=2$$

METHODOLOGY $2$:  APPLY STOLZ-CESARO 
Using the Stolz-Cesaro Theorem, we find that 
$$\begin{align}
\lim_{n\to \infty}\frac{\sum_{k=1}^n \frac1{\sqrt k}}{\sqrt n}&=\lim_{n\to \infty} \left(\frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}\right)\\\\
&=\lim_{n\to \infty}\left(1+\frac{\sqrt n}{\sqrt{n+1}}\right)\\\\
&=2
\end{align}$$
A: This sequence does converge. For every positive integer $n$, since $x \mapsto 1/\sqrt{x}$ is decreasing, we have
$$ \int_0^{n} \frac{dx}{\sqrt{x}} > \sum_{k=1}^n \frac{1}{\sqrt{k}} >\int_1^{n+1} \frac{dx}{\sqrt{x}} \\
2 \sqrt{n} > \sum_{k=1}^n \frac{1}{\sqrt{k}} >2 (\sqrt{n+1}-1) \\
2  >\frac{1}{\sqrt{n}}\sum_{k=1}^n \frac{1}{\sqrt{k}} >2\left(\sqrt{1+\frac{1}{n}}-\frac{1}{\sqrt{n}}\right) \to 2 $$
as $n \to \infty$.
