How do I evaluate $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}$? How do i evaluate this limit : $$\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+............+\frac{1}{\sqrt{n}}\right)$$ ?
Thank you for any help .
 A: \begin{align}
&\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{n}}\right) \\
= \ &\frac{1}{n}\left(\sqrt{\frac{n}{1}} + \sqrt{\frac{n}{2}} + \ldots + \sqrt{\frac{n}{n}}\right) \\
= \ &\frac{1}{n}\left(\frac{1}{\sqrt{\frac{1}{n}}} + \frac{1}{\sqrt{\frac{2}{n}}} + \ldots + \frac{1}{\sqrt{\frac{n}{n}}}\right) \\
\rightarrow \ &\int_0^1 \frac{1}{\sqrt{x}} \mathrm{d}x = 2
\end{align}
A: We bound $s_n=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}$ by integrals.  We get
$$\int_1^n \frac{dt}{\sqrt{t}}\le s_n\le 1+\int_1^{n-1} \frac{dt}{\sqrt{t}}.$$
Ealuate the integrals, divide by $\sqrt{n}$, and Squeeze.
A: Let  $a_n = 1 + \dfrac {1}{\sqrt{2}} + \cdots + \dfrac {1}{\sqrt{n}}$ and $b_n = \sqrt{n}$ for all $n\geq 1$. We have
$$\lim_{n\to \infty} \dfrac {a_{n+1} - a_n} {b_{n+1} - b_n} = \lim_{n\to \infty} \dfrac {1} {\sqrt{n+1}(\sqrt{n+1} - \sqrt{n})} = \lim_{n\to \infty} \dfrac {\sqrt{n+1} + \sqrt{n}} {\sqrt{n+1}} = 2,$$ 
so by the Stolz-Cesàro theorem we get $\lim_{n\to \infty} \dfrac {a_n} {b_n} = 2$
