The question is from Pui Ching Invitational Mathematics Competition (2019)
My attempt:
$x_{n+1} = \frac{5}{\sqrt{x_{n}+8}+\sqrt{x_{n}+3}} $
if $x_{n}> 1$，$x_{n+1}< 1$
if $x_{n}< 1$，$x_{n+1}> 1$
$x_{1}= 2019$, $x_{2}= 0.0555...$, $x_{odd}>1$, $x_{even}<1$,
let $x_{2k-1}= 1+α$, and $x_{2k}= 1-β$,
in the domain of $ 0<x<1 $, $\sqrt{9+x}-\sqrt{4+x} > 1-x $, therefore
$1-β = \sqrt{9+α}-\sqrt{4+α} > 1-α, α>β$
Similarly，$\sqrt{9-x}-\sqrt{4-x} < 1+x $, therefore
$1+α' = \sqrt{9-β}-\sqrt{4-β} < 1+β, α'<β$
the amplitudes of the $α$ & $β$ grossly decrease, and the sequence converges to $1$
$x_{1}= 2019$
$x_{2}= 0.0555...$
$x_{3}= 1.09.....$
$x_{4}= 0.9925...$
$x_{5}= 1.000619...$
$x_{6}= 0.999948...$
$x_{2019}\sim1$
Calculate the pair sums of odd term and even term since $x_{5}+x_{6}:$
$2014 < (x_{5}+x_{6})+(x_{7}+x_{8})+...+(x_{2017}+x_{2018})< (x_{5}+x_{6})*1007 < 2014.571…$
$4036.138 < x_{1}+x_{2}+...+x_{2018}+x_{2019} < 4036.709…$
I got the answer of $4036$
However, my method is somewhat awkward, and it should has a more elegant solution for the problem. Thank you for your help.