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?

We want to find the integer part of 
$\sum_{x=1}^{50} f(x)$
where
$f(x) = \sqrt{2x}-\sqrt{2x-1}$.
For decreasing $f(x)$, 
$$\int_a^{b+1} f(x) \; dx < \sum_{x=a}^b f(x) < \int_{a-1}^b f(x) \; dx$$
(It may help to make a sketch to see this.)
It's easier to work with the sum from $x=2$ to $50$ instead of $x=1$ to 50 because $f(x)$ is not defined when $x=0$.  We can adjust for the missing $x=1$ term later.
By elementary calculus, 
$$\int f(x) \; dx = \frac{1}{3} (2x)^{3/2} - \frac{1}{3} (2x-1)^{3/2} + C$$
so
$$\int_2^{51} f(x) \; dx < \sum_{x=2}^{50} f(x) < \int_{1}^{50} f(x) \; dx$$
yields
$$4.10 < \sum_{x=2}^{50} f(x) < 4.38$$
Now to adjust for the missing $x=1$ term.  Since $$\sum_{x=2}^{50} f(x) + \sqrt{2} - \sqrt{1} = \sum_{x=1}^{50} f(x)$$
we have
$$4.10 + \sqrt{2} - \sqrt{1} < \sum_{x=1}^{50} f(x) < 4.38 + \sqrt{2} - \sqrt{1}$$
which yields
$$4.51 < \sum_{x=1}^{50} f(x) < 4.79$$
so the integer part of the sum is $\boxed{4}$.
