About One Problem in Wildt 2018

Question

The W32 in the 28th József Wildt International Mathematical Competition is as follows:

$\text{If }x_{0}=1\text{ and }x_{n+1}^3+1=(x_{n}+1)^3\text{ for all }n\geq0\text{, then }$ $[x_{n}]\!=\mspace{-2mu}n\text{ for all }n\geq1\text{, when }\![\cdot]\mspace{-1mu}\text{ denote the integer part.}$

It has been solved in this post. Hence we know that $\lim\limits_{n\to\infty}\dfrac{x_{n}}{n}=1$ or $$x_n\sim{n}$$ as $n\to\infty$. But what can we know about $\mathtt{L}\mathop{:=}\lim\limits_{n\to\infty}(x_{n}-n)$? Since $\lceil{x_n}\rceil=n+1$ for $n\in\mathbb{N}$, the problem is only evaluating $\lim\limits_{n\to\infty}\langle{x_n}\rangle$ or determining $$x_n\sim n+\mathtt{L}$$ as $n\to\infty$, where numerical experiment shows that in fact $0.6<\mathtt{L}<0.7$.
However, can one express $\mathtt{L}$ in terms of some other well-known constants (if it exists)? Thanks.

Answer 1

It's probably not a well known constant, but we can study its asymptotes.

Write $y_n = x_n - n$, so that $(n + y_n)^3 + 1 = (n + y_{n - 1})^3$ holds for all $n \ge 1$.

As in the linked post, we can prove by induction that $y_n > \frac 1{n + 1}$.

Moreover, from $(n + y_n)^3 < (n + y_n)^3 + 1 = (n + y_{n - 1})^3$ we get $y_n < y_{n - 1}$. So the sequence $(y_n)_n$ decreases and is bounded below (by zero), hence converges to a limit $L$.

Writing $z_n = y_{n - 1} - y_n$, we have $$z_n = \frac 1{(n + y_{n - 1})^2 + (n + y_{n - 1})(n + y_n) + (n + y_n)^2}\tag 0$$ which is contained in the interval $[\frac 1{3(n + 1)^2}, \frac 1{3n^2}]$.

Thus we have $L = y_n + \sum_{k = n + 1}^\infty z_k$, hence $y_n = L - \sum_{k = n + 1}^\infty z_k = L - \frac 1{3n} + O(\frac 1{n^2})$ where we use the result $z_k \in [\frac 1{3(k + 1)^2}, \frac 1{3k^2}]$.

Putting this asymptote of $y_n$ back into $(0)$, we get $z_n = \frac 1{3n^2}(1 - \frac{2L}n + O(\frac 1{n^2}))$; we can then use this again with Euler-Maclaurin formula to get even higher asymptotes; etc. The procedure continues and gives rise to an asymptotic expansion of $y_n$ (and hence $x_n$).