Let $\{ y_k \}$ that satisfies $ y_k\le {2^k\over M}y_{k-1}^\beta$ , then $\lim_{k\to \infty}y_k=0$. Let be a sequence $\{ y_k \}^\infty _{k=0} \subset (0,\infty) $  that satisfies 
$$ y_k\le {2^k\over M}y_{k-1}^\beta    ,   $$ 
where  $k=1,2,...$, and   $\beta\gt 1$ , $M\gt0$.
Prove that if  $M\gt2^{\beta\over \beta-1}y_0^{\beta-1}$, then $\lim_{k\to \infty}y_k=0$.

Try to compare with
$$
ca^ky_k=u_k\le u_{k-1}^β= (ca^{k-1}y_{k-1})^β
\\\iff\\ 
y_k\le c^{β-1}a^{(k-1)β-k}y_{k-1}^β=(c^{β-1}a^{-β})(a^{β-1})^ky_{k-1}^β
$$
which is successful by identifying the relations $2=a^{β-1}$, $M^{-1}= a^{-β}c^{β-1}$, that is
$$
a=2^{\frac1{β-1}},\quad c=M^{-\frac1{β-1}}2^{\frac{β}{(β-1)^2}}
$$
It follows that
$$
u_n\le u_0^{β^n}
$$
which converges to zero for $0\le u_0<1$, i.e., 
$$
0\le M^{-\frac1{β-1}}2^{\frac{β}{(β-1)^2}}y_0<1
$$
which can be transformed into the given condition.

Since $a>1$ the sequence $y_k$ converges to zero even if $u_k$ is bounded, so that also equality in the condition is permissible.
