Make upper bounds for $\frac{1}{2^{n-k-1}}\;.\frac{1}{1-2^{\beta-(n-k)}}$. I want to make upper bounds for the following fraction respected to $\beta$ only (not $k$ or $n$).
\begin{align*}\frac{1}{2^{n-k-1}}\;.\frac{1}{1-2^{\beta-(n-k)}}
\end{align*}
where $\beta‎‎\simeq 0.7$, $0\leq k\leq n$, and ‎$k, ‎n\in‎\mathbb{N}‎$.
Since $0\leq k\leq n$ then we get $2^{n-k-1}\geq\frac{1}{2}$. So we have
\begin{align*}
\frac{1}{2^{n-k-1}}\leq2
\end{align*}
Now we should find upper bounds for $\frac{2}{1-2^{\beta-(n-k)}}$. here I tried as follows
\begin{align*}
n-k\geq0 &\Rightarrow \beta-(n-k)\leq \beta \\&\Rightarrow 1-2^{\beta-(n-k)}\geq1-2^{\beta}
\end{align*}
but I could not make as I mentioned upper bounds in this way, because $1-2^{\beta}\leq0$. Any hint can help me.  

For each integers  $0\le k\le n$ put 
$$f(n,k)=\frac{1}{2^{n-k-1}}\;.\frac{1}{1-2^{\beta-(n-k)}}=\frac 1{2^{n-k-1}-2^{\beta-1}}.$$
If $n=k$ then $f(n, k)$ is negative. If $n-k\ge 1$ then $f(n,k)\le \frac 1{1-2^{\beta-1}}$  and the equality is attained when $n-k=1$. Thus the required tight upper bound is $\frac 1{1-2^{\beta-1}}$.
