11G (Number Theory)

With

p_{- 1} = 1, p_{0} = a_{0}

q_{- 1} = 0, q_{0} = 1

and

p_{n} = a_{n} p_{n - 1} + p_{n - 2}

q_{n} = a_{n} q_{n - 1} + q_{n - 2}

, induction gives

p_{n} q_{n - 1} - p_{n - 1} q_{n} = (- 1)^{n + 1}

[a_{0}, \dots, a_{n}, β] = (p_{n} β + p_{n - 1}) / (q_{n} β + q_{n - 1})

; monotonicity in

β > 0

places it strictly between the adjacent convergents.

b

Apply the formula to the complete quotient and use the determinant identity to express

b_{n}

. Eliminating the complete quotient from the consecutive expressions yields the stated quadratic for

q_{n + 1} / q_{n}

. Taking the compatible signs for adjacent indices and using

c_{n} \leq b_{j}

gives

2 c_{n} a_{n + 1} \leq 2 b_{n} a_{n + 1} \leq 1 - 4 b_{n} b_{n + 1} + 1 - 4 b_{n - 1} b_{n} \leq 2 1 - 4 c_{n}^{2} .

If all but finitely many convergents had

b_{n} > 1/ 5

, then eventually

c_{n} > 1/ 5

. Part (b) would imply

c_{n} a_{n + 1} \leq 1 - 4 c_{n}^{2} < c_{n}

, impossible since

a_{n + 1} \geq 1

. Thus infinitely many

b_{n} \leq 1/ 5

, which is exactly

∣ θ - p_{n} / q_{n} ∣ \leq 1/ (5 q_{n}^{2})

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