Prove by the method of differences that
$\sum_{r=1}^{N} \frac{1}{\sqrt{r} + \sqrt{r+1}} = \sqrt{N+1} - 1.$

1. Rationalise the denominator of the general term.

$\frac{1}{\sqrt{r} + \sqrt{r+1}} = \frac{1}{\sqrt{r+1} + \sqrt{r}} \times \frac{\sqrt{r+1} - \sqrt{r}}{\sqrt{r+1} - \sqrt{r}}$
$= \frac{\sqrt{r+1} - \sqrt{r}}{(r+1) - r} = \sqrt{r+1} - \sqrt{r}$

2. Expand the sum to list the first few and last few terms, identifying the telescoping pattern.

$\sum_{r=1}^{N} (\sqrt{r+1} - \sqrt{r})$
Evaluating the terms from $r = 1$ to $r = N$:
When $r=1$: $\sqrt{2} - \sqrt{1}$
When $r=2$: $\sqrt{3} - \sqrt{2}$
When $r=3$: $\sqrt{4} - \sqrt{3}$
$\dots$
When $r=N-1$: $\sqrt{N} - \sqrt{N-1}$
When $r=N$: $\sqrt{N+1} - \sqrt{N}$

3. Cancel the intermediate terms to find the final simplified sum.

Summing the terms, all intermediate square roots cancel out:
$= \sqrt{N+1} - \sqrt{1}$
$= \sqrt{N+1} - 1$