How do I evaluate the following expression?

Question

How to evaluate the following expression:

$\displaystyle \frac{1}{\sqrt{2}+1}+ \frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} +\cdots +\frac{1}{\sqrt{9}+\sqrt{8}}$

Answer 1

Observe that:

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

Therefore,

$\dfrac1{\sqrt{2} + 1} = \sqrt{2} - 1.$

Similarly,

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

And therefore,

$\dfrac1{\sqrt{3} + \sqrt{2}} = \sqrt{3} - \sqrt{2}.$

We keep doing this until we reach,

$\dfrac1{\sqrt{8} + \sqrt{9}} = \sqrt{9} - \sqrt{8}.$

Now we add all the above equations to get:

$\displaystyle \frac{1}{\sqrt{2}+1}+ \frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} +\cdots +\frac{1}{\sqrt{9}+\sqrt{8}}.$

$=\displaystyle (\sqrt{2} - 1) + (\sqrt{3}-\sqrt{2}) + (\sqrt{4} - \sqrt{3}) +\cdots + (\sqrt{9}-\sqrt{8}).$

Notice that $+\sqrt{2}$ cancels with $-\sqrt{2}$. And this happens with all the numbers in the expression, except with $1$ and $\sqrt{9}$.

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

Answer 2

HINT:

$$\frac1{\sqrt{r+1}+\sqrt r}=\frac{\sqrt{r+1}-\sqrt r}{r+1-r}=\sqrt{r+1}-\sqrt r$$

Can you see the Telescoping Series?