We are asked to find the value of the following sum:

$S = \frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}}.$

Step 1: Rationalizing the denominators

Each term in the sum has the form $\frac{1}{\sqrt{n} + \sqrt{n+1}}$, where $n$ is an integer. To simplify this, we rationalize the denominator by multiplying both the numerator and the denominator by $\sqrt{n+1} - \sqrt{n}$.

For example, for the first term $\frac{1}{1 + \sqrt{2}}$, we multiply the numerator and denominator by $\sqrt{2} - 1$:

$\frac{1}{1 + \sqrt{2}} = \frac{1}{1 + \sqrt{2}} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1.$

Similarly, for the second term $\frac{1}{\sqrt{2} + \sqrt{3}}$, we multiply by $\sqrt{3} - \sqrt{2}$:

$\frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}.$

Step 2: Applying this to all terms

We can apply this rationalization to all terms in the sum. The general term will look like this:

$\frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}.$

So, each term in the sum becomes:

$\frac{1}{1 + \sqrt{2}} = \sqrt{2} - 1, \quad \frac{1}{\sqrt{2} + \sqrt{3}} = \sqrt{3} - \sqrt{2}, \quad \frac{1}{\sqrt{3} + \sqrt{4}} = \sqrt{4} - \sqrt{3}, \quad \dots$

Thus, the sum becomes:

$S = (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + (\sqrt{5} - \sqrt{4}) + (\sqrt{6} - \sqrt{5}) + (\sqrt{7} - \sqrt{6}) + (\sqrt{8} - \sqrt{7}) + (\sqrt{9} - \sqrt{8}).$

Step 3: Simplifying the sum

Notice that this is a telescoping series. Most terms cancel out with adjacent terms:

$S = -1 + \sqrt{9} = -1 + 3 = 2.$

Thus, the value of the sum is:

$\boxed{2}.$

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Would you like further explanation or any other details? Here are some related questions:

1. How does the telescoping property help simplify sums like this?
2. What happens if we have a different number of terms in the sum?
3. Can this method be applied to sums with different types of radicals?
4. How do we rationalize denominators in general when dealing with square roots?
5. What other types of sums can be simplified using similar techniques?

Tip: Always look for cancellation patterns (like telescoping) when summing terms that have similar structures.