Question

Find the first convergent to $\sqrt{101}$ that is correct to five places of decimals.

Answer 1

We can write $\sqrt{101}=10+\sqrt{101}-10=10+\frac{1}{\sqrt{101}+10}$

$\Rightarrow \sqrt{101}+10=20+\sqrt{101}-10=20+\frac{1}{\sqrt{101}+10}$

Therefore, the continued fraction is :-

$\sqrt{101}=10+\frac{1}{20+\frac{1}{20+\frac{1}{20+....}}}$

The convergents are :-

$\frac{10}{1},\frac{201}{20},\frac{4030}{401},.....$

The third convergent differs from $\sqrt{101}$ by $\frac{1}{{\left(401\right)}^2}$ and is therefore correct to $5$ places of decimal.