(Absolute) convergence of $\sum_{k=1}^\infty \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1}}$ We want to check if the following series converges (absolutely).
$$\sum_{k=1}^\infty \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1}}$$
This is what I have so far:
\begin{align}
\sum_{k=1}^\infty\frac{\sqrt{k+1}-\sqrt k}{\sqrt{k+1}}
&=\sum_{k=1}^\infty\frac{(\sqrt{k+1}-\sqrt k)(\sqrt{k+1}+\sqrt k)}{\sqrt{k+1}(\sqrt{k+1}+\sqrt k)}\\
&=\sum_{k=1}^\infty\frac1{\sqrt{k+1}(\sqrt{k+1}+\sqrt k)}\\
&= \sum_{k=1}^\infty \frac{1}{k+1+\sqrt{k^2+k}}
\end{align}
But how do we continue from here?

We have
$$\frac{1}{k+1+\sqrt{k^2+k}} > \frac{1}{k+1+\sqrt{k^2+2k+1}} = \frac{1}{2(k+1)} .$$
Thus the comparison test shows that $\sum_{k=1}^\infty\frac{\sqrt{k+1}-\sqrt k}{\sqrt{k+1}}$ diverges.
