I am reading the Wolfram page for 'convergents' (of continued fractions) and it states without proof the following inequality $$\frac{1}{(a_{n+1}+2)q_n^2}<|\alpha-\frac{p_n}{q_n}|<\frac{1}{a_{n+1}q_n^2}$$ Where $\alpha = [a_1, a_2, ...]$ as a continued fraction with convergents $\frac{p_n}{q_n}$. To obtain this, I presume they use the inequality $$\frac{1}{q_n q_{n+2}}<|\alpha-\frac{p_n}{q_n}|<\frac{1}{q_nq_{n+1}}$$ and then apply the relation $q_n = a_nq_{n-1}+q_{n-2}$. This gives the right hand inequality straight away. However I can't see how to derive the other inequality. In case you're wondering, I'm interested in this because I want to show that $$|\sqrt{2}-\frac{p}{q}|\ge\frac{1}{4q^2}$$ using the continued expansion for $\sqrt{2}$ so any hints for that would also be appreciated.


For the left hand inequality I think they use the following inequality: $$\left| \alpha- \frac{p_n}{q_n}\right| > \frac{1}{q_n(q_{n+1}+q_n)}$$ then: $$q_{n+1} = a_{n+1} q_n+q_{n-1} \leq (a_{n+1}+1) q_n$$ to obtain: $$q_n(q_n+1+q_n) \leq q_n((a_{n+1}+2)q_n)$$ which gives the inequality.

  • $\begingroup$ Thanks! That looks great. Can you give e a reference to the inequality you used at the beginning. I'm not sure How I would derive that. $\endgroup$ – Elie Bergman Apr 18 '18 at 10:09
  • $\begingroup$ I don't know any exact reference but the key ideas are the following: * Prove that $$\frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}}=\frac{(-1)^{n+1}}{q_n q_{n-1}}$$ * Then by monotonicity $$\left| \frac{p_n}{q_n}-\alpha \right|> \left|\frac{p_n}{q_n}-\frac{p_{n+2}}{q_{n+2}} \right|$$ * Notice that $$\frac{p_n}{q_n}-\frac{p_{n+2}}{q_{n+2}}=\frac{p_n}{q_n}-\frac {p_{n+1}}{q_{n+1}} +\frac {p_{n+1}}{q_{n+1}}-\frac{p_{n+2}}{q_{n+2}}$$ * Use this and the first formula to show: $$\left|\frac{p_n}{q_n}-\frac{p_{n+2}}{q_{n+2}} \right|=\frac{1}{q_n q_{n+1}}-\frac{1}{q_{n+1} q_{n+2}}$$ $\endgroup$ – Delta-u Apr 18 '18 at 11:37
  • $\begingroup$ And as $q_{n+2} \geq q_{n+1} + q_n$ you finally obtain: $$\left| \frac{p_n}{q_n}-\alpha \right|> \frac{1}{q_n q_{n+1}}-\frac{1}{q_{n+1} (q_n +q_{n+1})}=\frac{1}{q_n (q_n+q_{n+1})}$$ $\endgroup$ – Delta-u Apr 18 '18 at 11:43

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