Give me a five!

$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\dots+\frac{1}{\sqrt{99}+\sqrt{100}}=A$

$(\frac{1}{\sqrt{0}+\sqrt{1}}+\frac{1}{\sqrt{1}+\sqrt{2}})+(\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}})+\dots+(\frac{1}{\sqrt{98}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{100}})=B$

It is easy to see that

$\frac{1}{\sqrt{0}+\sqrt{1}}>\frac{1}{\sqrt{1}+\sqrt{2}}>\dots>\frac{1}{\sqrt{99}+\sqrt{100}}$

From these we get: $B>2A.$

By calculating we get:

$B=\sqrt{1}+(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\dots+(\sqrt{100}+\sqrt{99})=10$

So $A<\frac{B}{2}=5$ .

By the concavity of the $\sqrt{x}$ function (since $(\sqrt{x})'' = - \frac {1}{4x^{\frac {3}{2}}} < 0$ for $x \geq 0$ ) we have $\sqrt{2k-1} > \frac {\sqrt{2k} + \sqrt{2k-2}}{2}$ or $\sqrt{2k} - \sqrt{2k-1} < \sqrt{2k-1} - \sqrt{2k-2}$ for all $k \geq 1$ , so we can derive the following:

$2x = 2\displaystyle \sum_{k=1}^{50} \frac {1}{\sqrt{2k-1} + \sqrt{2k}} = 2\displaystyle \sum_{k=1}^{50} \frac {\sqrt{2k} - \sqrt{2k-1}}{(\sqrt{2k-1} + \sqrt{2k})(\sqrt{2k} - \sqrt{2k-1})} = 2\displaystyle \sum_{k=1}^{50} \frac {\sqrt{2k} - \sqrt{2k-1}}{2k - (2k-1)} =$

$= 2\displaystyle \sum_{k=1}^{50} \sqrt{2k} - \sqrt{2k-1} < \displaystyle \sum_{k=1}^{50} \sqrt{2k} - \sqrt{2k-1} + \displaystyle \sum_{k=1}^{50} \sqrt{2k-1} - \sqrt{2k-2} =$

$= \displaystyle \sum_{k=1}^{100} \sqrt{k} - \sqrt{k-1} = \sqrt{100} - \sqrt{0} = 10$

Hence $x < 5$ .