Find the sum $\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{99}+\sqrt{100}}$

Question

I would like to check I have this correct

Find the sum $$\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{99}+\sqrt{100}}$$ Hint: rationalise the denominators to get a 'telescoping' sum: a sum of terms in which many pairs add up to zero.

I rationalised the denominators to get a series like this: $$\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1} +...+\frac{\sqrt{99}-\sqrt{100}}{-1} $$ Which can be written: $$\sqrt{2}-\sqrt{1} + \sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}...+ \sqrt{99}-\sqrt{98} +\sqrt{100}-\sqrt{99}$$ Which is the telescoping sum the question talks about. Most of the terms drop out to leave $$-\sqrt{1} +\sqrt{100} = 9$$

Have I got this correct?

Answer 1

you could also do it by induction

conjecture: $\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{n-1}+\sqrt{n}} = \sqrt{n}-1$ For $n \in [2,3,4 ...]$

For $n=2$:

$$\frac{1}{\sqrt{1}+\sqrt{2}} = \sqrt{2}-1$$

For $n+1$: $$\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{n-1}+\sqrt{n}} + \frac{1}{\sqrt{n}+\sqrt{n+1}}$$

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

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

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

$$Q.E.D.$$