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

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    $\begingroup$ Please share what you have tried. (Don't answer it in the comments, add it to the question) $\endgroup$
    – S.C.B.
    Sep 24, 2016 at 14:49
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    $\begingroup$ I would write it as the integral of $x^{-1/2}$, evaluate it, and hope to show the error between the integral and the sum is small enough not to cross an integer. Presumably whoever posed the problem made sure that works. $\endgroup$ Sep 24, 2016 at 14:51
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    $\begingroup$ You might consider using the notation $\lfloor S_n \rfloor$ for the `floor' function. $\endgroup$
    – Myridium
    Sep 24, 2016 at 15:46

3 Answers 3

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

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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:

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

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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 $18$.

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