How to find $x$ such that $\frac{1}{2^{p-1}} -\frac{1}{a^p}>x$? Let $a, p\in\mathbb{R}$ be such that $a\ge 2$ and $p>1$. Consider the quantity
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}.$$
Since $a\ge 2$, it seems clear to me that
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}>0.$$
My question is: it is possible to find a positive real number $x$ (not depending on $a$) such that
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}>x?$$
Could someone please help me in finding that?
Thank you in advance!

$$\frac{1}{2^{p-1}} -\frac{1}{a^p}\geq \frac{1}{2^{p-1}} -\frac{1}{2^p}$$
So if you want strict inequality, we simply have:
$$\frac{1}{2^{p-1}} -\frac{1}{a^p} > \frac{1}{2}\left(\frac{1}{2^{p-1}} -\frac{1}{2^p}\right)$$
