(I've tried to represent the fractions and roots as clearly as possible within the limits of ASCII text).
First off, I only looked at the first fraction:
1 ÷ (√1+√2)
Because we don't like roots in the denominator, I'm going to remove them by multiplying the difference of two squares, like so:
1 ÷ (√1+√2) × (√1-√2) ÷ (√1-√2)
Although I'm multiplying the fraction by another, I haven't actually changed the size of the number as (√1-√2) ÷ (√1-√2) cancels to 1.
The result of this leaves me with:
(√1-√2) ÷ (√1+√2)(√1-√2)
Which can be simplified like so:
(√1-√2) ÷ ((√1)²+(√1)(√2)-(√1)(√2)-(√2)²)
(√1-√2) ÷ (1-2)
(√1-√2) ÷ -1
(√2-√1) ÷ 1
(√2-√1)
Thus, 1 ÷ (√1+√2) is equal to (√2-√1).
From this solution, I find that it would be very useful if all of the fractions in Alison's question were able to be left just as surds rather than fractions.
I can check for this algebraically, like so:
1 ÷ (√a+√(a+1))
1 ÷ (√a+√(a+1)) × (√a-√(a+1)) ÷ (√a-√(a+1))
(√a-√(a+1)) ÷ (√a+√(a+1))(√a-√(a+1))
(√a-√(a+1)) ÷ ((√a)²-(√(a+1))²)
(√a-√(a+1)) ÷ (a-(a+1))
(√a-√(a+1)) ÷ -1
√(a+1)-√a
Now that I can see this applies to all fractions following this rule, I'll look to see how much this helps me.
The first thing I noticed is how the largest number is always the positive. Because the numbers in the square roots increase each time, this means that the biggest number will always be the smallest number in at least one other fraction, given that it is not the first or last.
Another way of wording it is that after the first fraction, I will be left with √2 - √1, and after the second fraction I am left with √3 - √2. What I noticed is that the √2 from the first result will cancel out with the -√2 from the second result, leaving us with just -√1.
Using this method, we can follow through this 'cancelling out' up until 1 ÷ (√99+√100), where the -√99 cancels with the √99 in the previous equation, and we are left with √100.
Taking our -√1 from the first equation we did, we are left with:
√100 - √1
Which quite clearly cancels to:
10 - 1
and,
9
I found this to be quite an interesting problem, with a very satisfying answer once found.