Solving the Square Root Series Sum

The problem asks us to find the sum of the following series:

$$ S = \frac 1 {\sqrt 1 + \sqrt 2} + \frac 1 {\sqrt 2 + \sqrt 3} + \frac 1 {\sqrt 3 + \sqrt 4} + \dots + \frac 1 {\sqrt {80} + \sqrt {81}} $$

This can be written in summation notation as:

$$ S = \sum_{n=1}^{80} \frac{1}{\sqrt{n} + \sqrt{n+1}} $$

Simplifying Terms Using Rationalization

To simplify the general term $\frac{1}{\sqrt{n} + \sqrt{n+1}}$, we can rationalize the denominator. We multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{n+1} - \sqrt{n}$:

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

Now, we simplify the expression:

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

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

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

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

Evaluating the Series Using Telescoping Sum

Now we can rewrite the series using the simplified terms:

$$ S = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \dots + (\sqrt{80} - \sqrt{79}) + (\sqrt{81} - \sqrt{80}) $$

This is a telescoping series. Notice that the positive term of each pair cancels out the negative term of the next pair:

• -√1 remains
• +√2 cancels with -√2
• +√3 cancels with -√3
• ...
• +√80 cancels with -√80
• +√81 remains

After cancellation, only the first negative term and the last positive term remain:

$$ S = \sqrt{81} - \sqrt{1} $$

Final Calculation

We know that $\sqrt{81} = 9$ and $\sqrt{1} = 1$. Therefore, the sum is:

$$ S = 9 - 1 = 8 $$

The sum of the given expression is 8.