Sum of square roots...........

Question

$$ \mbox{If}\quad S = 1 + \,\sqrt{\,\frac{1}{2}\,}\, + \,\sqrt{\,\frac{1}{3}\,}\, + \,\sqrt{\,\frac{1}{4}\,}\, + \,\sqrt{\,\frac{1}{5}\,}\, + \cdots + \,\sqrt{\,\frac{1}{100}\,}\,\,, $$ 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 $S$.

Answer 1

Hint: $$S=\sum_{k=1}^{100}\frac{1}{\sqrt{k}}$$ $$\frac{1}{\sqrt{k+1}+\sqrt{k}}\lt \frac{1}{2\sqrt{k}}\lt \frac{1}{\sqrt{k+\frac{1}{2}}+\sqrt{k-\frac{1}{2}}}$$ $$\implies \sqrt{k+1} -\sqrt{k} \lt \frac{1}{2\sqrt{k}}\lt \sqrt{k+\frac12}-\sqrt{k-\frac{1}{2}}$$

Answer 2

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:

$$\int_1^{100} \sqrt{\frac{1}{x}}\ \text{d}x = 2\sqrt{x}\to 2\sqrt{100} - 2\sqrt{1} = 18$$

Answer 3

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 $18$.