let $$a_1=1, a_{n+1} = a_n + \frac{1}{a_n^2}$$, and it seems $$\lim_{n\to\infty} (a_n^3-3n-\log n) = \text{C}(constant) $$
I use computer program to verify that $$C = 1.13525585...$$,and expecting someone here could shed some light on this value with a closed form expression or more real digits with some efficient numerical method, thanks.
(this is a Chinese mathematics forum, FYI: http://bbs.emath.ac.cn/thread-8790-1-1.html)
Hint: Try cubing both sides of the recurrence to get some sense of how $a_n^3$ grows