Question

Find the first convergent to $\sqrt{15}$ that is correct to five places of decimals.

Answer 1

We can write $\sqrt{15}=3+\sqrt{15}-3$

$=3+\frac{6}{\sqrt{15}+3}$

$\Rightarrow \frac{\sqrt{15}+3}{6}=1+\frac{\sqrt{15}-3}{6}=1+\frac{1}{\sqrt{15}+3}$

$\Rightarrow \sqrt{15}+3=6+\sqrt{15}-3=6+\frac{6}{\sqrt{15}+3}$

Therefore, $\sqrt{15}=3+\frac{1}{1+\frac{1}{6+\frac{1}{1+\frac{1}{6+...}}}}$

The convergents are:-

$\frac{3}{1},\frac{4}{1},\frac{27}{7},\frac{31}{8},\frac{213}{55},\frac{244}{63},\frac{1677}{433}.....$

The og differs from $\sqrt{15}$ by less than $\frac{1}{{\left(433\right)}^2}$ and hence it is correct to $5$ places of decimal.