Evaluate the sum \[\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dotsb + \frac{1}{\sqrt{15} + \sqrt{16}}.\]

(You might want to use a calculator to get an estimate of the answer, but in order to get the exact answer you will have to do it by hand!)

We can think of \(3 - 2\) as a difference of two squares, since \(3 - 2 = \left(\sqrt{3}\right)^2 - \left(\sqrt{2}\right)^2\).

Let the sum we’re interested in be \(S\), so \[S = \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dotsb + \frac{1}{\sqrt{15} + \sqrt{16}}.\]

Then \[S = \frac{\sqrt{2} - \sqrt{1}}{2 - 1} + \frac{\sqrt{3} - \sqrt{2}}{3 - 2} + \frac{\sqrt{4} - \sqrt{3}}{4 - 3} + \dotsb + \frac{\sqrt{16} - \sqrt{15}}{16 - 15}.\]

We can simplify the denominators very nicely, to get \[S = (-\sqrt{1} + \sqrt{2}) + (-\sqrt{2} + \sqrt{3}) + (-\sqrt{3} + \sqrt{4}) + \dotsb + (-\sqrt{15} + \sqrt{16}).\]

The sums on the right-hand side now telescope (lots of adjacent terms cancel), so we are just left with the first and last terms.

So \(S = \sqrt{16} - \sqrt{1} = 4 - 1 = 3\).

Can you find a similar sum that evaluates to \(5\)?

We could construct a similar expression with sum \(5\) by adding on extra terms up to \(\frac{1}{\sqrt{35} + \sqrt{36}}\).

There are other possibilities too: if we construct a similar sum that evaluates to \(8\), then the difference between this and \(S\) will be \(3\). So we can do something similar even if we don’t start with \(\frac{1}{\sqrt{1} + \sqrt{2}}\).

Can you find a similar sum that evaluates to a number that is not an integer?

We can construct many similar expressions that don’t evaluate to integers. One example would be \(T = S - \frac{1}{\sqrt{15} + \sqrt{16}}\) (so just miss out the last term of \(S\)). If we rationalise the denominators then we still get a telescoping sum, but this time \(T = \sqrt{15} - \sqrt{1} = \sqrt{15} - 1\), which is not an integer.