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 \[\frac{1}{q_{k+1}q_{k+2}} \lt \frac{1}{q_kq_{k+1}}\]