Original problem by reddit. Solution and extension by Adrian Hernandez Vega

Original Problem

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Consider:

$$ S = \frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{9}+\sqrt{11}}+...\frac{1}{\sqrt{9997}+\sqrt{9999}} $$

Is \(S\) less than or greater than \(24\)?

Solution 1: Integration Solution

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Observe that the sequences under the radicals are formed by \(4n-3\) and \(4n-1\). We set the bounds to be from \(1\) to \(2500\).

$$ S = \sum_{n=1}^{2500}\frac{1}{\sqrt{4n-3}+\sqrt{4n-1}} $$

Rationalizing the denominator:

$$ \begin{aligned} & = \sum_{n=1}^{2500} \frac{1}{\sqrt{4n-3}+\sqrt{4n-1}}\cdot\frac{\sqrt{4n-3}-\sqrt{4n-1}}{\sqrt{4n-3}-\sqrt{4n-1}} \\ &= \sum_{n=1}^{2500}\frac{\sqrt{4n-1}-\sqrt{4n-3}}{2} \end{aligned} $$

It follows that:

$$ \sum_{n=1}^{2500}\frac{\sqrt{4n-1}-\sqrt{4n-3}}{2}>\int_{1}^{2500}\frac{\sqrt{4n-1}-\sqrt{4n-3}}{2}dn $$

We will split the right side into indefinite integrals, for simplicity. Substituting \(u_1 = 4n-3\) and \(u_2 = 4n-1\):

$$ \begin{aligned} & \int \frac{\sqrt{4n-1}-\sqrt{4n-3}}{2}dn \\ & =\frac{1}{2}\left(\int \sqrt{4n-1}dn- \int \sqrt{4n-3}dn\right) \\ & =\frac{1}{2}\left(\frac{1}{4}\int\sqrt{u_{1}}du_{1}-\frac{1}{4}\int\sqrt{u_{2}}du_{2}\right) \end{aligned} $$

Using the standard power rule, \(\int \sqrt{x} dx = \frac{2}{3} x^{\frac{3}{2}} + C\)

$$ \begin{aligned} & =\frac{1}{2}\left(\frac{1}{4}\left(\frac{2}{3}\left(4n-1\right)^{\frac{3}{2}}\right)-\frac{1}{4}\left(\frac{2}{3}\left(4n-3\right)^{\frac{3}{2}}\right)\right) \\ & =\frac{1}{12}\left(\left(4n-1\right)^{\frac{3}{2}}-\left(4n-3\right)^{\frac{3}{2}}\right) \\ \end{aligned} $$

Substituting the bounds \(1\) to \(2500\), the integral expression is roughly equal to \(24.647820496\), thus proving \(S>24\).

$$ \sum_{n=1}^{2500}\frac{\sqrt{4n-1}-\sqrt{4n-3}}{2}>24.647820496 $$

Solution 2: Direct Comparison

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Take the previous setup, but instead notice that \(\frac{1}{\sqrt{4n-3} + \sqrt{4n-1}} > \frac{1}{\sqrt{4n-1} + \sqrt{4n-1}}\) in our domain.

$$ \begin{aligned} S &= \sum_{n=1}^{2500}\frac{1}{\sqrt{4n-3}+\sqrt{4n-1}} > \sum_{n=1}^{2500}\frac{1}{\sqrt{4n-1}+\sqrt{4n-1}} \\ &\implies S > \sum_{n=1}^{2500}\frac{1}{2\sqrt{4n-1}} > \int_{1}^{2500} \frac{1}{2\sqrt{4n-1}} dn \end{aligned} $$

Evaluating the right-side integral and letting \(u=4n-1\):

$$ \begin{aligned} S &> \int_{1}^{2500} \frac{1}{2\sqrt{4n-1}} dn \\ &> \int_{3}^{9999} \frac{1}{8\sqrt{u}} du \\ &> \frac{\sqrt{9999}}{4} - \frac{\sqrt{3}}{4} \\ &> 24.5657372669 \end{aligned} $$