Let $\{x_{n}\}_{n=0}^{\infty}$ be decreasing sequence of non-negative reals. Suppose that there exist constants $a, s>0$ and $b>1$ such that $$x_{n+1}\leq ab^{n}x_{n}^{1+s}$$ and $$x_{0}\leq a^{-1/s}b^{-1/s^{2}}.$$ Is it true that then $$\lim_{n\to \infty}x_{n}=0?$$

Any help is appreciated!


1 Answer 1


By induction on $n$, we check that $$x_n\le a^{-1/s}b^{-1/s^2-n/s}$$ for all integers $n\ge0$.

Now the desired result immediately follows.

The condition that $x_n$ is decreasing in $n$ was not needed or used.

  • $\begingroup$ Thank you very much! Could the induction basis be something else here or is it related to the above condition? $\endgroup$
    – Shaq155
    May 3, 2022 at 6:35
  • $\begingroup$ @Shaq155 : The initial condition, on $x_0$, is very important here. I think the upper bound on $x_0$ cannot be improved, but I have not checked that. I am wondering where this problem came from. $\endgroup$ May 3, 2022 at 13:19
  • $\begingroup$ Do you remember what that paper was? $\endgroup$ May 3, 2022 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.