How to prove $|a_{2n}-a_{n}|<\frac{10}{27}$ Let sequence $$a_{1}=1,a_{n+1}=\dfrac{2}{2a_{n}+1}$$
show that
$$\left|a_{2n}-a_{n}\right|<\dfrac{10}{27}$$
and the constant  $\frac{10}{27}$ A smaller number instead
$$a_{n+1}+\dfrac{1+\sqrt{17}}{4}=2(1+\sqrt{17})\cdot\dfrac{a_{n}+\frac{1+\sqrt{17}}{4}}{4(2a_{n}+1)}\tag{1}$$
$$a_{n+1}-\dfrac{-1+\sqrt{17}}{4}=-2(-1+\sqrt{17})\cdot\dfrac{a_{n}-\frac{-1+\sqrt{17}}{4}}{4(2a_{n}+1)}\tag{2}$$
$\dfrac{(1)}{(2)}$,that
$$\dfrac{a_{n+1}+\dfrac{1+\sqrt{17}}{4}}{a_{n+1}-\dfrac{-1+\sqrt{17}}{4}}=\dfrac{1+\sqrt{17}}{1-\sqrt{17}}\cdot\left(\dfrac{a_{n}+\dfrac{1+\sqrt{17}}{4}}{a_{n}-\dfrac{-1+\sqrt{17}}{4}}\right)$$
so we I get ugly even more can't solve this problem.can you some recommended or solve this?

The function $f(x)=\frac{1}{x+1/2}$ maps $[2/3,1]$ to $[2/3,6/7] \subset [2/3,1]$, since $1/3 < 10/27$ the result follows.
