Find the integral part of $\sum_{i=2}^{10000}\frac1{\sqrt{i}}$ $$A = \frac1{\sqrt{2}}+\frac1{\sqrt{3}}+\cdots+\frac{1}{\sqrt{10000}}$$
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.
 A: We start by noting that, $$\frac{1}{\sqrt{k}}=\frac2{\sqrt{k}+\sqrt{k}}\lt\frac2{\sqrt{k}+\sqrt{k-1}}$$
So we have, $$\frac{1}{\sqrt{k}}\lt \frac{2}{\sqrt{k}+\sqrt{k-1}} = 2(\sqrt{k} - \sqrt{k-1})$$
Thus we have $$S=\sum_{i=2}^{10000}\frac{1}{\sqrt{k}} \lt \sum_{i=2}^{10000}2(\sqrt{k} - \sqrt{k-1}) = 2(\sqrt{10000}-\sqrt1) = 198$$
$$\color{red}{A\lt198}\tag{1}$$
Also, 
$$\frac{1}{\sqrt{k}}=\frac2{\sqrt{k}+\sqrt{k}}\gt\frac2{\sqrt{k}+\sqrt{k+1}}$$
And so, 
$$\frac{1}{\sqrt{k}}\gt \frac{2}{\sqrt{k}+\sqrt{k+1}} = 2(\sqrt{k+1} - \sqrt{k})$$
And therefore,
$$S=\sum_{i=2}^{10000}\frac{1}{\sqrt{k}} \gt \sum_{i=2}^{10000}2(\sqrt{k+1} - \sqrt{k}) = 2(\sqrt{10001}-\sqrt2) \gt 197$$
$$\color{red}{A\gt197}\tag{2}$$
Combining $(1)$ and $(2)$
$$197\lt A \lt 198$$
$$\implies \lfloor A\rfloor = 197$$
A: $$
\begin{align}
197 &< 2\left(\sqrt{10001}-\sqrt{2}\right) \\
&= \int_2^{10001} \frac{dx}{\sqrt{x}} \\
&< \color{blue}{\sum_{k=2}^{10000} \frac{1}{\sqrt{k}}} \\
&< \int_1^{10000} \frac{dx}{\sqrt{x}} \\
&= 198.
\end{align}
$$
