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!



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




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The condition that $x\_n$ is decreasing in $n$ was not needed or used.


