Can you make it larger?

We know that $S = \displaystyle \sum_{k=2}^{10000} {\dfrac{1}{\sqrt{k}}} \approx I = \int_2^{10000} {\dfrac{1}{\sqrt{x}} dx} = 197.1715729$ .

In the graphs above, $S$ is represented by total area of the vertical bars and $I$ is the area under the curve $\frac{1}{\sqrt{x}}$ .

Looking at the left graph, the first bar of $S$ centers at $x=1$ . If we remove the area of half of the first bar $(\frac{1}{\sqrt{2}}\times \frac{1}{2})$ and half of last bar $(\frac{1}{\sqrt{10000}}\times \frac{1}{2})$ (not shown in the graph), we note that:

$S - \frac{1}{2}\times \frac{1}{\sqrt{2}} - \frac{1}{2}\times \frac{1}{\sqrt{10000}} > I$

This is because the small area $a$ of each bar is larger than the area $b$ that follows.

$\Rightarrow S > I + \frac{1}{2}\left( \frac{1}{\sqrt{2}} + \frac{1}{100} \right) = 197.5301263$

Looking at the right graph, where the first bar of $S$ centers at $x=0.5$ , we note that $S$ without the first bar is smaller than $I$ :

$\Rightarrow S - \frac {1}{\sqrt{2}} < I$

This is because all $a=0$ .

$\Rightarrow S < \frac {1}{\sqrt{2}} + I = 197.8786797$

Therefore,

$197.5301263 < S < 197.8786797 \quad \Rightarrow \lfloor S \rfloor = \boxed{197}$