Alternating summation and subtraction of square roots I encountered a problem, to find the integer part of: $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{3} + \sqrt{4}} +...+\frac{1}{\sqrt{99} + \sqrt{100}}$$.
I multiplied the conjugate of each denominator.  Meaning, for $\frac{1}{\sqrt{a} + \sqrt{b}}$, I multiply $\sqrt{a} - \sqrt{b}$.  I get $\sqrt{100} - \sqrt{99} + \sqrt{98} - \sqrt{97} + ... + \sqrt{2} - \sqrt{1}$.  I am stuck here.  How do I simplify this?  Or is it that my method is wrong?

$$\frac {1}{\sqrt n +\sqrt {n+1}} = \sqrt {n+1} - \sqrt {n}$$
Therefore the  dsire sum is $ A-B$ where,  
$$A= \sqrt 100+\sqrt 98 +....+\sqrt 2 = 338.047...$$ and 
$$B=\sqrt 99+\sqrt 97 +....+\sqrt 1 = 333.415....$$
Thus [A-B]=4
