How to get $\sqrt {k} + \frac{1}{\sqrt{k+1}}$ in the form $\frac{\sqrt{k^2} + 1}{\sqrt{k+1}}$? I was wondering if it is possible to get $\sqrt {k} + \dfrac{1}{\sqrt{k+1}}$ in the form $\dfrac{\sqrt{k^2} + 1}{\sqrt{k+1}}$, and if so, how?
I ask this, because I'm following this answer, and I get lost at how they arrive at $\dfrac{\sqrt{k^2} + 1}{\sqrt{k+1}}$.
Thanks.
EDIT: To further clarify what I'm not understanding, please read below:
I don't understand how the author goes from the line:
$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} +...+ \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} > \sqrt{k+1}$
to the line:
$\frac{\sqrt{k^2 + k} + 1}{\sqrt{k + 1}} > \frac{\sqrt{k^2} + 1}{\sqrt{k+1}}$
I would really appreciate it if someone could lead me through to that line and explain exactly what is being done on each line.
Thanks again.

It cannot hold obviously. When $k=\color{red}1$,  $\sqrt {k} + \frac{1}{\sqrt{k+1}}=\color{red}{1+\frac{\sqrt2}{2}}$, however, $\frac{\sqrt{k^2} + 1}{\sqrt{k+1}}=\color{red}{\sqrt2}.$
