## Accuracy of Convergents of Infinite Continued Fractions

Theorem (Accuracy of Convergents)
If
$\frac{p_1}{q_1}, \; \frac{p_2}{q_2}, \; \frac{p_3}{q_3},,,$
is a sequence of convergents for a continued fraction
$x= [ a_1, \; a_2,... ]$
, then
$\| x - \frac{p_{k+1}}{q_{k+1}} \| \lt \frac{1}{q_{k+1}q_{k+2}} \| \le \frac{1}{2q_kq_{k+1}} \| \lt \| x - \frac{p_k}{q_k} \|$
.
Proof
Let
$x= [ a_1, \; a_2,... ]$
. The Continued Fraction Algorithm yields the system of equations
$x= [ q_1, \; x_2 ] = [ a_1, \_2, \; x_3 ] = ...= [a_1, \; a_2,..., a_n, \; z_{n+1} ] =...$

and, using The Value of a Finite Continued Fraction Theorem then the Properties of Convergents of Finite Continued Fractions
\begin{equation} \begin{aligned} \| x - \frac{p_n}{q_n} \| &= [ a_1, \; a_2,..., \; a_n, \; x_{n+1} ] - \frac{p_n}{q_n} ] \\ &= \| \frac{x_{n+1}p_n+p_{n-1}}{x_{n+1}q_n+q_{n-1}}\| \\ &= \| \frac{p_{n-1} q_n-p_nq_{n-1}}{q_n(x_{n+1}q_n+q_{n-1})} \| \\ &= \frac{1}{q_n(x_{n+1}q_n+q_{n-1})} \end{aligned} \end{equation}

$x_{n+1}=[ a_{n+1}, \; a_{n+2},...$
so
$a_{n+1} \lt x_{n+1} \lt a_{n+1}+1$
, and
$\| x- \frac{p_n}{q_n} \| \lt \frac{1}{q_n(a_{n+1}q_n+q_{n-1})} = \frac{1}{q_nq_{n+1}}$
which gives the left hand inequality by taking
$n=k+1$
.
Also
\begin{equation} \begin{aligned} \| x- \frac{p_n}{q_n} \| & \gt \frac{1} {q_n((a_{n+1}+1)q_n+q_{n-1})} \\ &= \frac{1}{q_n(a_{n+1}q_n+q_{n-1})+q^2_n} \\ &= \frac{1}{q_n(q_{n+1}+q_n)} \\ & \gt \frac{1}{q_n(q_{n+1}+q_{n+1})} \\ &= \frac{1}{2q_nq_{n+1}} \end{aligned} \end{equation}

which gives the left hand inequality when
$n=k+1$
.
which gives the right hand inequality by taking
$n=k$
.
For the middle inequality note that
$q_{k+2}q_{k+1}q_{k+1}+q_k \ge q_k+q_k = 2q_k$
so
$\frac{1}{q_{k+1}q_{k+2}} \lt \frac{1}{q_kq_{k+1}}$ 