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.

You should observe that your series telescopes, i.e.:
$$\sum_{k=0}^n (\sqrt{k+1} - \sqrt{k}) = (\sqrt{1} -\sqrt{0}) + (\sqrt{2} -\sqrt{1}) +\cdots + (\sqrt{n}-\sqrt{n-1}) +(\sqrt{n+1}-\sqrt{n}) = \sqrt{n+1}-1\; ,$$
and therefore:
$$\sum_{k=0}^\infty (\sqrt{k+1} - \sqrt{k}) = \lim_{n\to \infty} \sum_{k=0}^n (\sqrt{k+1} - \sqrt{k}) = \lim_{n\to \infty} \sqrt{n+1}-1 = \infty\; .$$
