Timeline for Why is the following recurrent sequence convergent?

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Sep 10, 2021 at 18:33	comment	added	Iosif Pinelis		@mlk : The problem is that $x_n$ may initially grow very fast. E.g., for $a = 1, b = 10, c = 100, d = 0;$ we have $(x_0,\dots,x_{10})\approx(1, 10, 1560, 106655, 4.2327910^6, 1.1092710^8, 2.0765610^9, 2.9305210^{10}, 3.2437510^{11}, 2.9035210^{12}, 2.15372*10^{13})$. However, I think at some point a subtle transition to a steady regime occurs, and I don't know how to capture this transition.
Sep 10, 2021 at 18:18	comment	added	mlk		@IosifPinelis True, I underestimated the square there. So only the "bounded -> convergent" part still holds. One can probably extract more terms to get a higher exponent $n^a$ in front of that term, but it might not be enough. Now I wonder if there might not be an unbounded sequence for large intial values.
Sep 10, 2021 at 14:21	comment	added	Iosif Pinelis		Alas, I do not think this will work. Indeed, the difference inequality $a_n -a_{n-1}\le\frac{c_1}{n^2} (a_{n-1}+a_{n-2}) + \frac{c_2}{n^3} a_{n-1}^2$ will hold if e.g. $a_n=(n+1)^2$, $c_1>0$, and $c_2=3$.
Sep 10, 2021 at 13:53	comment	added	A. PI		Thank you for this answer, but how will you use the "discrete-Gronwall" argument ! The last term in the r.h.s has a square. This was the mistake that killed the answer here math.stackexchange.com/questions/4221679/… f
Sep 10, 2021 at 12:52	history	answered	mlk	CC BY-SA 4.0