Accuracy of Convergents of Continued Fraction Expansion of Irrational Number

Theorem
Let $$x$$ be an irrational number.

Let $$\left \langle {C_n}\right \rangle$$ be the sequence of convergents of $$x$$.

Let $$p_1, p_2, p_3, \ldots$$ and $$q_1, q_2, q_3, \ldots$$ be its numerators and denominators.

Then:
 * $$\forall k \ge 1: \left|{x - \frac {p_{k+1}} {q_{k+1}}}\right| \le \frac 1 {q_{k+1} q_{k+2}} \le \frac 1 {2 q_k q_{k+1}} < \left|{x - \frac {p_k} {q_k}}\right|$$.

Corollary
$$\forall k \ge 1: \frac 1 {q_k q_{k+1}} > \left|{x - \frac {p_k} {q_k}}\right| > \frac 1 {2 q_k q_{k+1}}$$

Proof
Let $$x$$ have an SICF of $$\left[{a_1, a_2, a_3, \ldots}\right]$$.

The Continued Fraction Algorithm gives the following system of equations:

$$ $$ $$ $$ $$

and

$$ $$ $$ $$

Now $$x_{n+1} = \left[{a_{n+1}, a_{n+2}, a_{n+3}, \ldots}\right]$$ from the Continued Fraction Algorithm.

So $$a_{n+1} < x_{n+1} < a_{n+1} + 1$$.

Therefore:
 * $$\left|{x - \frac {p_n} {q_n}}\right| < \frac 1 {q_n \left({a_{n+1} q_n + q_{n-1}}\right)} = \frac 1 {q_n q_{n+1}}$$.

This gives the LHS of the inequality when $$n = k+1$$.

We also have:
 * $$\left|{x - \frac {p_n} {q_n}}\right| > \frac 1 {q_n \left({\left({a_{n+1} + 1}\right) q_n + q_{n-1}}\right)}$$

$$ $$ $$ $$ $$

This gives the RHS of the inequality when $$n = k$$.

For the middle inequality, note that:
 * $$q_{k+2} = a_{k+2} q_{k+1} + q_k > q_k + q_k = 2 q_k$$

So:
 * $$\frac 1 {q_{k+1} q_{k+2}} \le \frac 1 {2 q_k q_{k+1}}$$

Proof of Corollary
Immediate.

Comment
The left hand side of the inequality gives an indication of how close each convergent gets to its true value.

The right hand side gives a bound that limits its accuracy.