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!

$$
 f(a) = \frac{1}{2^{p-1}} -\frac{1}{a^p}
$$
is increasing in $a$, therefore is
$$
f(a) \ge f(2) = \frac{1}{2^{p-1}} -\frac{1}{2^p}= \frac{1}{2^p}
$$
for $a \ge 2$.
