How do I evaluate the following expression? How to evaluate the following expression:  
$\displaystyle \frac{1}{\sqrt{2}+1}+ \frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} +\cdots +\frac{1}{\sqrt{9}+\sqrt{8}}$

Observe that:
$(\sqrt{2} + 1)(\sqrt{2} - 1) = (\sqrt{2})^2 - 1^2 = 2 - 1 = 1.$ 
Therefore,
$\dfrac1{\sqrt{2} + 1} = \sqrt{2} - 1.$
Similarly,
$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 2 - 1 = 1.$ 
And therefore,
$\dfrac1{\sqrt{3} + \sqrt{2}} = \sqrt{3} - \sqrt{2}.$
We keep doing this until we reach,
$\dfrac1{\sqrt{8} + \sqrt{9}} = \sqrt{9} - \sqrt{8}.$
Now we add all the above equations to get:
$\displaystyle \frac{1}{\sqrt{2}+1}+ \frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} +\cdots +\frac{1}{\sqrt{9}+\sqrt{8}}.$
$=\displaystyle (\sqrt{2} - 1) + (\sqrt{3}-\sqrt{2}) + (\sqrt{4} - \sqrt{3}) +\cdots + (\sqrt{9}-\sqrt{8}).$
Notice that $+\sqrt{2}$ cancels with $-\sqrt{2}$. And this happens with all the numbers in the expression, except with $1$ and $\sqrt{9}$.
$= \sqrt{9} - 1 = 3 - 1 = 2.$
