Answer :
To solve the expression
[tex]\[ \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}} \][/tex]
we will simplify each fraction and observe if there is a pattern or if they can be summed directly.
First, consider the general form of each term: \(\frac{1}{\sqrt{n} + \sqrt{n+1}}\).
We will rationalize the denominator. For the general term:
[tex]\[ \frac{1}{\sqrt{n} + \sqrt{n+1}} \cdot \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1})^2 - (\sqrt{n})^2} \][/tex]
Since \((\sqrt{n+1})^2 - (\sqrt{n})^2 = (n + 1) - n = 1\), the fraction becomes:
[tex]\[ \frac{\sqrt{n+1} - \sqrt{n}}{1} = \sqrt{n+1} - \sqrt{n} \][/tex]
Now, applying this to each fraction in the sum:
[tex]\[ \begin{align*} \frac{1}{1+\sqrt{2}} &= \sqrt{2} - 1 \\ \frac{1}{\sqrt{2}+\sqrt{3}} &= \sqrt{3} - \sqrt{2} \\ \frac{1}{\sqrt{3}+\sqrt{4}} &= 2 - \sqrt{3} \\ \frac{1}{\sqrt{4}+\sqrt{5}} &= \sqrt{5} - 2 \\ \frac{1}{\sqrt{5}+\sqrt{6}} &= \sqrt{6} - \sqrt{5} \\ \frac{1}{\sqrt{6}+\sqrt{7}} &= \sqrt{7} - \sqrt{6} \\ \frac{1}{\sqrt{7}+\sqrt{8}} &= 2\sqrt{2} - \sqrt{7} \\ \frac{1}{\sqrt{8}+\sqrt{9}} &= 3 - 2\sqrt{2} \end{align*} \][/tex]
Next, sum all these expressions:
[tex]\[ (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3}) + (\sqrt{5} - 2) + (\sqrt{6} - \sqrt{5}) + (\sqrt{7} - \sqrt{6}) + (2\sqrt{2} - \sqrt{7}) + (3 - 2\sqrt{2}) \][/tex]
Notice that this is a telescoping series, where most terms will cancel out with subsequent terms. Let's write out the full sum to see the cancellation clearly and simplify:
[tex]\[ \begin{align*} (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3}) + (\sqrt{5} - 2) + (\sqrt{6} - \sqrt{5}) + (\sqrt{7} - \sqrt{6}) + (2\sqrt{2} - \sqrt{7}) + (3 - 2\sqrt{2}) = \\ \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3} + \sqrt{5} - 2 + \sqrt{6} - \sqrt{5} + \sqrt{7} - \sqrt{6} + 2\sqrt{2} - \sqrt{7} + 3 - 2\sqrt{2} \end{align*} \][/tex]
Here, we observe that \(\sqrt{2}\) cancels with \(-\sqrt{2}\), \(\sqrt{3}\) cancels with \(-\sqrt{3}\), \(\sqrt{5}\) cancels with \(-\sqrt{5}\), \(\sqrt{6}\) cancels with \(-\sqrt{6}\), \(\sqrt{7}\) cancels with \(-\sqrt{7}\), and \(2\sqrt{2}\) cancels with \(-2\sqrt{2}\).
What remains is:
[tex]\[ -1 + 3 = 2 \][/tex]
Thus, the sum of the given series of fractions is:
[tex]\[ \boxed{2} \][/tex]
[tex]\[ \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}} \][/tex]
we will simplify each fraction and observe if there is a pattern or if they can be summed directly.
First, consider the general form of each term: \(\frac{1}{\sqrt{n} + \sqrt{n+1}}\).
We will rationalize the denominator. For the general term:
[tex]\[ \frac{1}{\sqrt{n} + \sqrt{n+1}} \cdot \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1})^2 - (\sqrt{n})^2} \][/tex]
Since \((\sqrt{n+1})^2 - (\sqrt{n})^2 = (n + 1) - n = 1\), the fraction becomes:
[tex]\[ \frac{\sqrt{n+1} - \sqrt{n}}{1} = \sqrt{n+1} - \sqrt{n} \][/tex]
Now, applying this to each fraction in the sum:
[tex]\[ \begin{align*} \frac{1}{1+\sqrt{2}} &= \sqrt{2} - 1 \\ \frac{1}{\sqrt{2}+\sqrt{3}} &= \sqrt{3} - \sqrt{2} \\ \frac{1}{\sqrt{3}+\sqrt{4}} &= 2 - \sqrt{3} \\ \frac{1}{\sqrt{4}+\sqrt{5}} &= \sqrt{5} - 2 \\ \frac{1}{\sqrt{5}+\sqrt{6}} &= \sqrt{6} - \sqrt{5} \\ \frac{1}{\sqrt{6}+\sqrt{7}} &= \sqrt{7} - \sqrt{6} \\ \frac{1}{\sqrt{7}+\sqrt{8}} &= 2\sqrt{2} - \sqrt{7} \\ \frac{1}{\sqrt{8}+\sqrt{9}} &= 3 - 2\sqrt{2} \end{align*} \][/tex]
Next, sum all these expressions:
[tex]\[ (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3}) + (\sqrt{5} - 2) + (\sqrt{6} - \sqrt{5}) + (\sqrt{7} - \sqrt{6}) + (2\sqrt{2} - \sqrt{7}) + (3 - 2\sqrt{2}) \][/tex]
Notice that this is a telescoping series, where most terms will cancel out with subsequent terms. Let's write out the full sum to see the cancellation clearly and simplify:
[tex]\[ \begin{align*} (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3}) + (\sqrt{5} - 2) + (\sqrt{6} - \sqrt{5}) + (\sqrt{7} - \sqrt{6}) + (2\sqrt{2} - \sqrt{7}) + (3 - 2\sqrt{2}) = \\ \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3} + \sqrt{5} - 2 + \sqrt{6} - \sqrt{5} + \sqrt{7} - \sqrt{6} + 2\sqrt{2} - \sqrt{7} + 3 - 2\sqrt{2} \end{align*} \][/tex]
Here, we observe that \(\sqrt{2}\) cancels with \(-\sqrt{2}\), \(\sqrt{3}\) cancels with \(-\sqrt{3}\), \(\sqrt{5}\) cancels with \(-\sqrt{5}\), \(\sqrt{6}\) cancels with \(-\sqrt{6}\), \(\sqrt{7}\) cancels with \(-\sqrt{7}\), and \(2\sqrt{2}\) cancels with \(-2\sqrt{2}\).
What remains is:
[tex]\[ -1 + 3 = 2 \][/tex]
Thus, the sum of the given series of fractions is:
[tex]\[ \boxed{2} \][/tex]
