(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?
 A: Note that
$$
\forall k \in \Bbb N,\quad \sqrt{k+1}-\sqrt{k} = \sqrt{k+1}\left(1-\sqrt{1-\frac{1}{k+1}}\right)
$$
and since $\sqrt{1+x} \underset{x\to 0}{=} 1 +\frac{1}{2}x + o(x)$, it follows that the general term in the summation satisfies
$$
\frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1}} \underset{k\to\infty}{=} \frac{1}{2(k+1)} + o\left(\frac{1}{k+1}\right).
$$
The general term in the summation is thus equivalent to $\frac{1}{2(k+1)}$, from which we deduce that the sum is divergent.
A: 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.
A: 
\begin{align}
S=\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}

Next, we have:
\begin{align}
\frac{1}{k+1+\sqrt{k^2+k}}>\frac{1}{k+k+\sqrt{k^2+k^2}}=\frac{1}{2k+\sqrt{2}k}>\frac{1}{4k}
\end{align}
\begin{align}
S>\sum_{k=1}^\infty\frac{1}{4k}
\end{align}
So, it is divergent.
