Given a sequence $\{x_n\}$ of positive real numbers converging to $0$, can we find a sequence of the form $\left\{\frac{1}{2^n}\right\}$ such that there exists $m,n_0\in\mathbb{N}$ with $$x_n^m\leq\frac{1}{2^n}\qquad\text{for all }\ n\geq n_0$$
Intuitively, it seems right (and sometimes wrong)! My approach of reasoning: -
Since $x_n\longrightarrow0$, $x_n<1$ for $n\geq n_0$. Therefore, one can make $x_n$ small enough by taking its $m$th power so that $$x_n^m\leq\frac{1}{2^n}\qquad\text{for all } n\geq n_1$$ where $n_1\geq n_0$.
I would be grateful if someone could clear my doubts on this. Thanks!
As @Kelenner has pointed out, if we consider the sequence $\left\{\frac{1}{n}\right\}$, then for a fixed $m$, $\frac{n^m}{2^n}\longrightarrow0$ as $n\rightarrow\infty$, i.e.,
$$\frac{1}{n^m}>\frac{1}{2^n},\ \ \ \ \ n\geq n_0 $$
for sufficiently large $n_0$ and hence the wrong conjecture.