finding a minorant to $(\sqrt{k+1} - \sqrt{k})$ Need help finding a minorant to
$(\sqrt{k+1} - \sqrt{k})$ which allows me to show that the series $\sum_{k=1}^\infty (\sqrt{k+1} - \sqrt{k})$ is divergent.

Since
$$\sum_{k=1}^n (\sqrt{k+1} - \sqrt{k})
=\sum_{k=1}^n ((\sqrt{k+1} - \sqrt{k})\frac{\sqrt{k+1} + \sqrt{k}}{\sqrt{k+1} + \sqrt{k}})
$$
$$
=\sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}}
\geq 2\sum_{k=1}^n \frac{1}{\sqrt{k+1}}
\geq 2\sum_{k=1}^n \frac{1}{k+1},
$$
then the series does not converge, but the telescope argument is much simpler.
