Given a natural number $n$ and a real number $d_{n},$ for the least $d_{n+1}$ so that $32d_{n}^{5}\!\left ( d_{n}+ 2 \right )\!=$ $$= 2\left ( 5d_{n}^{4}+ 8d_{n}^{2}+ 8d_{n}+ 8 \right )d_{n+ 1}^{3}+ 4d_{n}\left ( 5d_{n}^{4}+ 11d_{n}^{3}- 2d_{n}^{2}+ 14d_{n}- 4 \right )d_{n+ 1}^{2}$$ $+ 4d_{n}^{3}\left ( 5d_{n}^{3}+ 2d_{n}^{2}+ 24d_{n}- 16 \right )d_{n+ 1}.$ Prove that if $d_{2}= \frac{8}{5},$ then $\left \{ d_{n} \right \}_{n> 1}$ is rational and $$\lim\frac{n^{2}}{\ln n}\left ( \frac{2}{n}- d_{n} \right )= 2$$ Source: AoPS/@Ji_Chen (still unsolved)

I'm into the related one here https://Artofproblemsolving.com/community/c6h318726p1713923 _I also see $d_{2}= \!\frac{8}{5}.$ Maybe they are same so I used discriminant to define the concrete value for $d_{n}$ but unsuccessfully, I need to the help.


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