Find the integral part of $\sum_{i=2}^{10000}\frac1{\sqrt{i}}$

Question

$$A = \frac1{\sqrt{2}}+\frac1{\sqrt{3}}+\cdots+\frac{1}{\sqrt{10000}}$$

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.

Answer 1

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 $(2)$

$$197\lt A \lt 198$$

$$\implies \lfloor A\rfloor = 197$$

Answer 2

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