How does $ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$ reduce to $ 1 - \frac{1}{\sqrt{n+2}}\;$? $$ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$$
Reduces to:
$$ 1 - \frac{1}{\sqrt{n+2}} $$
I have no clue how. What is the exact trick here and how can I practice this? I can't properly google the problem.

HINT
Here is a different approach for the sake of curiosity (as suggested by Paul in the comments).
Precisely, we shall multiply the numerator and the denominator by the conjugate of the denominator:
\begin{align*}
\frac{1}{(n+2)\sqrt{n+1} + (n+1)\sqrt{n+2}} & = \frac{(n+2)\sqrt{n+1} - (n+1)\sqrt{n+2}}{(n + 2)^{2}(n+1) - (n+1)^{2}(n+2)}\\\\
& = \frac{(n+2)\sqrt{n+1} - (n+1)\sqrt{n+2}}{(n+1)(n+2)}\\
& = \frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n+2}}
\end{align*}
Can you take it from here?
