Reciprocal of Square Roots Sum

This solution is similar to Dhira's, and fills in the gap to show that $S < 25$ .

Define:

$S = \frac 1{\sqrt {1} + \sqrt {3} } + \frac 1{\sqrt {5} + \sqrt {7} } + \frac 1{\sqrt {9} + \sqrt {11} } + \cdots + \frac 1{\sqrt {9997} + \sqrt {9999}}$ $Q = \frac 1{\sqrt {3} + \sqrt {5} } + \frac 1{\sqrt {7} + \sqrt {9} } + \frac 1{\sqrt {11} + \sqrt {13} } + \cdots + \frac 1{\sqrt {9999} + \sqrt {10001}}$

By comparing termwise, we see that $S > Q$ .
By comparing a term of Q with the subsequent term of S, we see that $S - \frac{1}{ \sqrt{ 1} + \sqrt{3} } < Q$ .

Hence, we conclude that $2S - \frac{ \sqrt{3} - \sqrt{1} } { 2} < S + Q < 2S$ .

$\begin{aligned} S + Q & = \frac 1{\sqrt {1} + \sqrt {3} } + \frac 1{\sqrt {3} + \sqrt {5} } + \frac 1{\sqrt {5} + \sqrt {7} } + \cdots + \frac 1{\sqrt {9999} + \sqrt {10001}}\\ \\ & = \frac 1{\sqrt {1} + \sqrt {3} }.\frac {\sqrt {3} - \sqrt {1}}{\sqrt {3} - \sqrt {1}} + \frac 1{\sqrt {3} + \sqrt {5} }.\frac {\sqrt {5} - \sqrt {3}}{\sqrt {5} - \sqrt {3}} + \frac 1{\sqrt {5} + \sqrt {7} }.\frac {\sqrt {7} - \sqrt {5}}{\sqrt {7} - \sqrt {5}} + \cdots + \frac 1{\sqrt {9999} + \sqrt {10001}}.\frac {\sqrt {10001} - \sqrt {9999}}{\sqrt {10001} - \sqrt {9999}} \\ \\ & = \frac {\sqrt {3} - \sqrt {1}}2 + \frac {\sqrt {5} - \sqrt {3}}2 + \frac {\sqrt {7} - \sqrt {5}}2 +\cdots + \frac {\sqrt {10001} - \sqrt {9999}}2 \\ \\ & = - \frac {\sqrt {1}}2 + (\frac {\sqrt {3}}2 - \frac {\sqrt {3}}2) + (\frac {\sqrt {5}}2 - \frac {\sqrt {5}}2) + \cdots +(\frac {\sqrt {9999}}2 - \frac {\sqrt {9999}}2) + \frac {\sqrt {10001}}2 \\ \\ & = \frac {\sqrt {10001} - \sqrt 1}2 \\ \\ \end{aligned}$

Hence, our orignal inequality is states:

$2 S - \frac { \sqrt{3} - \sqrt{1 } } { 2 } < \frac { \sqrt{ 10001 } - \sqrt{ 1} } { 2 } < 2 S$

Evaluating the radicals, we get that

$49.503 < 2S < 49.868$

Hence, $\lfloor S \rfloor = 24$ .

Let $S = \frac 1{\sqrt {1} + \sqrt {3} } + \frac 1{\sqrt {5} + \sqrt {7} } + \frac 1{\sqrt {9} + \sqrt {11} } + \cdots + \frac 1{\sqrt {9997} + \sqrt {9999}}$ and $Q = \frac 1{\sqrt {3} + \sqrt {5} } + \frac 1{\sqrt {7} + \sqrt {9} } + \frac 1{\sqrt {11} + \sqrt {13} } + \cdots + \frac 1{\sqrt {9999} + \sqrt {10001}}$ We can see that S>Q $\begin{aligned} S + Q & = \frac 1{\sqrt {1} + \sqrt {3} } + \frac 1{\sqrt {3} + \sqrt {5} } + \frac 1{\sqrt {5} + \sqrt {7} } + \cdots + \frac 1{\sqrt {9999} + \sqrt {10001}}\\ \\ & = \frac 1{\sqrt {1} + \sqrt {3} }.\frac {\sqrt {3} - \sqrt {1}}{\sqrt {3} - \sqrt {1}} + \frac 1{\sqrt {3} + \sqrt {5} }.\frac {\sqrt {5} - \sqrt {3}}{\sqrt {5} - \sqrt {3}} + \frac 1{\sqrt {5} + \sqrt {7} }.\frac {\sqrt {7} - \sqrt {5}}{\sqrt {7} - \sqrt {5}} + \cdots + \frac 1{\sqrt {9999} + \sqrt {10001}}.\frac {\sqrt {10001} - \sqrt {9999}}{\sqrt {10001} - \sqrt {9999}} \\ \\ & = \frac {\sqrt {3} - \sqrt {1}}2 + \frac {\sqrt {5} - \sqrt {3}}2 + \frac {\sqrt {7} - \sqrt {5}}2 +\cdots + \frac {\sqrt {10001} - \sqrt {9999}}2 \\ \\ & = - \frac {\sqrt {1}}2 + (\frac {\sqrt {3}}2 - \frac {\sqrt {3}}2) + (\frac {\sqrt {5}}2 - \frac {\sqrt {5}}2) + \cdots +(\frac {\sqrt {9999}}2 - \frac {\sqrt {9999}}2) + \frac {\sqrt {10001}}2 \\ \\ & = \frac {\sqrt {10001} - \sqrt 1}2 \\ \\ & \approx 49.5 \\ \\ 2S & > 49.5 \\ \\ S & > \frac {49.5}2 > 24 \\ \\ \lfloor {S} \rfloor & = 24 \end{aligned}$