TITLE: How do I evaluate $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}$?
QUESTION [6 upvotes]: 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 .

REPLY [5 votes]: \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}

REPLY [3 votes]: 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.