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.
 A: Do you need a minorant? Just consider the partial sums. By telescoping you see that $\sum\limits_{k=1}^n \sqrt{k+1} - \sqrt k = \sqrt{n+1} - 1$, which diverges.
A: 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:
$$\sum_{k=0}^\infty (\sqrt{k+1} - \sqrt{k}) = \lim_{n\to \infty} \sum_{k=0}^n (\sqrt{k+1} - \sqrt{k}) = \lim_{n\to \infty} \sqrt{n+1}-1 = \infty\; .$$
A: 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.
A: If you do want a minorant, then you can use
$$
\sqrt{k+1}-\sqrt k = \int_k^{k+1} \frac1{2\sqrt x} dx > \int_k^{k+1} \frac1{2\sqrt {k+1}} dx = \frac1{2\sqrt {k+1}}.
$$
A: Note that
$$
\sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}>\frac{1}{2\sqrt{k+1}}
$$
