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Existence of Limit

Express the f...

Question

Express the following surds as continued fractions, and find the sixth convergent to $\sqrt{8}$ .

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Solution

$\sqrt{8} = 2 \sqrt{2}$

$= 2 [1 + (\sqrt{2} - 1)]$

$= 2 [1 + \frac{1}{1 + \sqrt{2}}]$

$= 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 1 + \frac{1}{1 + 1 + \frac{1}{1 + \sqrt{2}}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$

$= 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 1 + \frac{1}{2 + \frac{1}{1 + \sqrt{2}}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$

$= 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 + \frac{1}{2 + \frac{1}{1 + (1 + \frac{1}{1 + \sqrt{2}})}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$= 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{\sqrt{2}}}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$= 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{\sqrt{2}} . . .}}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$⟹ \sqrt{2} = [1, 2, 2, 2, . . .]$

Convergents of $\sqrt{2} ⟶ 1, \frac{3}{2}, \frac{7}{5}, \frac{17}{12}, \frac{41}{39}, \frac{99}{70}, . . .$

$∴ 2 \sqrt{2} ⟶ 2, 3, \frac{14}{5}, \frac{17}{6}, \frac{82}{39}, \frac{99}{25}, . . .$

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