Show that the sum of the series is greater than 24 $$\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{9}+\sqrt{11}} +\cdots+\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
I see that
$\frac{1}{\sqrt{3}+\sqrt{1}}\cdot \frac{\sqrt{3}-\sqrt{1}}{\sqrt{3}-\sqrt{1}}=\frac{\sqrt{3}-\sqrt{1}}{2}$
In each term, the denominator is 2 so the series becomes $\frac{1}{2}\sum^{2499}_{n=0}[\sqrt{4n+3}-\sqrt{4n+1}]$
This was not far from being telescopic. So let us make this telescopic.
$$4S> \frac{2}{\sqrt{1}+\sqrt{3}}+ \frac{2}{\sqrt{3}+\sqrt{5}}+\frac{2}{\sqrt{5}+\sqrt{7}} +...+ \frac{2}{\sqrt{9997}+\sqrt{9999}}+ \frac{2}{\sqrt{9999}+\sqrt{10001}}$$ $$ =\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\ldots+\sqrt{9999}-\sqrt{9997}+\sqrt{10001}-\sqrt{9999} $$ $$ =\sqrt{10001}-1>99 $$
The simplest way I can think of is to write it as $\frac12(\sum(\sqrt{4n+3} - \sum(\sqrt{4n-1}).$ You can approximate each sum by the integral, and since the function $\sqrt{x}$ is monotonic (and concave up) you can estimate the difference very closely.
A second (maybe even simpler) argument is to use your form, and note that by the mean value theorem each term equals $\frac{1}2 \frac{1}{\sqrt{x}}$ for some $x$ between $4n+1$ and $4n+3.$
