Proving "$\forall_{\epsilon>0} \exists_{N \in \mathbb{N}} \forall_{n \geq N}: x^{n} < \epsilon$" for $0<x<1$.

Question

I'm trying to prove a rather simple analysis statement but I think I'm overseeing something. For $0<x<1$ I need to prove

$$\forall_{\epsilon>0} \exists_{N \in \mathbb{N}} \forall_{n \geq N}: x^{n} < \epsilon.$$

I now have the following:

We use theorem 1.5 which states

\begin{equation}\label{theorem_1.5} \forall_{\epsilon>0}, \exists_{N \in \mathbb{N}} : \frac{1}{N} < \epsilon. \end{equation}

Also we make use of the fact that if $0<x<1$ it follows that

\begin{equation}\label{equation_1} (N+1)x^{N}<\frac{1}{1-x}. \end{equation}

Suppose $0<x<1$ and let $\epsilon>0$ be given. Now let $\epsilon'=\frac{1}{1-x}\epsilon$. Then from equation 1, equation 2 and the fact that $N+1>N$ it follows that \begin{equation} \begin{split} (N+1)x^{N} &< \frac{1}{1-x},\\ x^{N} &< \frac{1}{1-x}\frac{1}{N+1} < \frac{1}{1-x}\frac{1}{N}\\ x^{n} &< \frac{1}{1-x} \epsilon\\ x^{n} &< \epsilon'. \end{split} \end{equation}

My gut feeling says there is something wrong with my reasoning. It might be the fact that I predetermine $\epsilon'$, but I'm not sure.

Any thoughts would be appreciated.

Answer 1

Put $y=1/x$, then $y>1$, hence $y=1+z$ for some $z>0$. Then we get, with Bernoulli:

$$\frac{1}{x^n}=y^n=(1+z)^n \ge 1+nz >nz.$$

Hence

$$ x^n < \frac{1}{nz}.$$

Can you proceed ?