Cauchy's convergence test excercise (sin(n) / 2^n): I can't understand one place in the proof.

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In my textbook, when proving that

$${x_n} = \frac{\sin1}{2} + \frac{\sin2}{2^2} + \dots + \frac{\sin n}{2^n}$$

is convergent with help of Cauchy's convergence test,

it's required that

$$0<\epsilon<1$$

on the following step of the proof:

$$\vert {x_{n+p} - x_n \vert}<\frac{1}{2^n}<\epsilon \Rightarrow 2^n>\frac{1}{\epsilon} \Rightarrow n\ln2 > -\ln\epsilon \Rightarrow n>-\frac{\ln\epsilon}{\ln2}$$

Why do we need epsilon to be smaller than 1 and greater than 0?

Thank you a lot.