Question

Express the following surds as continued fractions, and find the fourth convergent to $\sqrt{1+\frac{1}{a}}$.

Answer 1

We can write $\frac{\sqrt{a^2+a}}{a}=1+\frac{\sqrt{a^2+a}-a}{a}$

$=1+\frac{1}{\sqrt{a^2+a}+a}$

$\sqrt{a^2+a}+a=2a+\sqrt{a^2+a}-a=2a+\frac{a}{\sqrt{a^2+a}+a}$

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

$\sqrt{1+\frac{1}{a}}=1+\frac{1}{2a}+\frac{1}{2}+\frac{1}{2a}+\frac{1}{2}+.....$

And the convergents are:-

$\frac{1}{1},\frac{2a+1}{2a},\frac{4a+3}{4a+1},\frac{8a^2+8a+1}{8a^2+4a}$