Cardinality of a set including fractions

Question

Attempt:

Let $G = \{x > 0 : \lim_{n→∞} \mathsf{frac}((n!)x) = 0\} .$

For each $f ∈ 2^ω$,

define $x_{f} = \sum_{n\ge1} \frac{f(n)}{n!}$

Put $W=\{x_{f} :f∈2^ω\}$.

I don't know how to proceed from here.

Answer 1

For any $f:(\Bbb Z^+\to \{0,1\}$ let $x_f=\sum_{m=2}^{\infty}f(m)/m!.$

If $f\ne g$ then $x_f\ne x_g.$ Proof: Consider $m_0=\min\{m:f(m)\ne g(m)\}$ and WLOG suppose $f(m_0)=1$ and $g(m_0)=0$. Then $m_0\ge 1$ so $x_f-x_g\ge \frac {1}{m_0!}-\sum_{m=1+m_0}^{\infty}\frac {1}{m!}>0.$

If $ n\in \Bbb Z^+$ then $$\text {frac}(n!x_f)=\sum_{m=n+1}^{\infty}n!f(m)/m!\le \sum_{m=n+1}^{\infty}n!/m!<\sum_{m=n+1}^{\infty}\frac {1}{(n+1)^{m-n}}=\frac 1 n.$$