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:
 * $\forall k \ge 1: \left\vert{x - \dfrac {p_{k+1}} {q_{k+1}} }\right\vert < \dfrac 1 {q_{k+1} q_{k+2}} \le \dfrac 1 {2 q_k q_{k+1}} < \left\vert{x - \dfrac {p_k} {q_k} }\right\vert$

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

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

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

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

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.