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?

For convenience, let $p=\frac{1+\sqrt{17}}{4}$ and $q=\frac{-1+\sqrt{17}}{4}$
Let $b_{n}=\frac{a_{n}+p}{a_{n}-q}$, then your last equation can be rewritten as follows:
$b_{n+1}=\frac{p}{q}b_{n}$
Since the above is a geometric sequence,  we have $b_{n}=b_{1}r^{n-1}$, where $r=\frac{p}{q}$.
Unsubstitting, we have $\frac{a_{n}+p}{a_{n}-q}=b_{1}r^{n-1}$.
So $a_{n}=\frac{qb_{1}r^{n-1}+p}{b_{1}r^{n-1}-1}$
I think you can derive your desired inequality using extra conditions such as $a_{1}=1$. 
