TITLE: Find the integral part of $\sum_{i=2}^{10000}\frac1{\sqrt{i}}$
QUESTION [6 upvotes]: $$A = \frac1{\sqrt{2}}+\frac1{\sqrt{3}}+\cdots+\frac{1}{\sqrt{10000}}$$
Find $\lfloor A\rfloor$ where $\lfloor x\rfloor$ is the greatest integer less than, or equal to $x$
I got stuck on this, so when I finally did it, I decided to post it here. And of course, I am always looking for alternatives, so keep answering.

REPLY [12 votes]: $$
\begin{align}
197 &< 2\left(\sqrt{10001}-\sqrt{2}\right) \\
&= \int_2^{10001} \frac{dx}{\sqrt{x}} \\
&< \color{blue}{\sum_{k=2}^{10000} \frac{1}{\sqrt{k}}} \\
&< \int_1^{10000} \frac{dx}{\sqrt{x}} \\
&= 198.
\end{align}
$$