# Sequence of reals such that $x_{n+1}\leq ab^{n}x_{n}^{1+s}$ converges to $0$?

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

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.

• Thank you very much! Could the induction basis be something else here or is it related to the above condition? May 3, 2022 at 6:35
• @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. May 3, 2022 at 13:19
• Do you remember what that paper was? May 3, 2022 at 15:54