Sum of square roots........... $$
\mbox{If}\quad S =
1 + \,\sqrt{\,\frac{1}{2}\,}\, + \,\sqrt{\,\frac{1}{3}\,}\, +
\,\sqrt{\,\frac{1}{4}\,}\, +
\,\sqrt{\,\frac{1}{5}\,}\, + \cdots + \,\sqrt{\,\frac{1}{100}\,}\,\,,
$$
then what is the value of $\left\lfloor\,S\,\right\rfloor$ ?.
Here $\left\lfloor\,S\,\right\rfloor$ is the greatest integer function which is less than or equal to $S$.
 A: Hint:
$$S=\sum_{k=1}^{100}\frac{1}{\sqrt{k}}$$
$$\frac{1}{\sqrt{k+1}+\sqrt{k}}\lt \frac{1}{2\sqrt{k}}\lt \frac{1}{\sqrt{k+\frac{1}{2}}+\sqrt{k-\frac{1}{2}}}$$
$$\implies \sqrt{k+1} -\sqrt{k} \lt \frac{1}{2\sqrt{k}}\lt \sqrt{k+\frac12}-\sqrt{k-\frac{1}{2}}$$
A: Just for the records of a pure mathematical curiosity, without proofs, the sum is everything but trivial. Indeed we have:
$$\sum_{k = 1}^{N}\sqrt{\frac{1}{k}} = \zeta \left(\frac{1}{2}\right)-\zeta \left(\frac{1}{2},N+1\right)$$
Where the first special function is the Riemann Zeta Function, and the second is the Hurwitz Zeta Function.
In your case:
$$\sum_{k = 1}^{100}\sqrt{\frac{1}{k}} = \zeta \left(\frac{1}{2}\right)-\zeta \left(\frac{1}{2},101\right)\approx 18.5896(...)$$
The integral test
To get a very good approximation in the continuum case, you can compute the integral:
$$\int_1^{100} \sqrt{\frac{1}{x}}\ \text{d}x = 2\sqrt{x}\to 2\sqrt{100} - 2\sqrt{1} = 18$$
A: The Euler-Maclaurin Summation Formula yields
$$\begin{align}
\sum_{k=1}^{100}k^{-1/2}&=1 +\int_1^{100} x^{-1/2}\,dx+\frac12\left(\frac{1}{\sqrt{100}}-1\right)+\frac1{24} \left(1-\frac{1}{(100)^{3/2}}\right)+R_2 \\\\
&=18.591625+R_2
\end{align}$$
where 
$$\begin{align}
\left|R_2\right|&\le \frac{2\zeta(2)}{(2\pi)^2}\int_1^{100} \left|\frac{d^2 x^{-1/2}}{dx^2}\right|\,dx\\\\
&= \frac{2\zeta(2)}{(2\pi)^2}\,\frac12 \left(1-\frac{1}{(100)^{3/2}}\right)\\\\
&=0.041625
\end{align}$$
Therefore, the integer part of the sum of interest is indeed $18$.
