summation - 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.

Answer

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 $(2)$

$$197\lt A \lt 198$$

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