Question

Express the following surds as continued fractions, and find the fourth convergent $\sqrt{a^2-\frac{a}{n}}$.

Answer 1

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

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

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

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

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

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

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

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

And the convergents are

$\frac{a-1}{1},\frac{a}{1},\frac{2an-a-1}{2n-1},\frac{2an-1}{2n}$