For $\mathrm{\mid }r\mathrm{\mid }<1$ the sequence $r^k$ is null, and for $\mathrm{\mid }r\mathrm{\mid }>1$ the sequence $\mathrm{\mid }r{\mathrm{\mid }}^k$ diverges to $+\mathrm{\infty }$

First, $\mathrm{\mid }{r}^k\mathrm{\mid }=\mathrm{\mid }r{\mathrm{\mid }}^k$ for every $k\in \mathbb{N}$, by induction: at $k=0$ both sides are $\mathrm{\mid }1\mathrm{\mid }=1$, and if $\mathrm{\mid }{r}^k\mathrm{\mid }=\mathrm{\mid }r{\mathrm{\mid }}^k$ then $\mathrm{\mid }{r}^{k+1}\mathrm{\mid }=\mathrm{\mid }{r}^{k}r\mathrm{\mid }=\mathrm{\mid }{r}^k\mathrm{\mid }\mathrm{\mid }r\mathrm{\mid }=\mathrm{\mid }r{\mathrm{\mid }}^k\mathrm{\mid }r\mathrm{\mid }=\mathrm{\mid }r{\mathrm{\mid }}^{k+1}$.

Case zero. Assume $r=0$.

Case small. Assume $0<\mathrm{\mid }r\mathrm{\mid }<1$.

Case large. Assume $\mathrm{\mid }r\mathrm{\mid }>1$.

In case zero, $r^k=0$ for every $k\ge 1$: indeed $r^1=r^{0}r=1\cdot 0=0$, and if $r^k=0$ then $r^{k+1}=r^{k}r=0$, so induction gives the claim from $k=1$ on.

In case small, put $s:=1\mathrm{/}\mathrm{\mid }r\mathrm{\mid }$, which is defined since $\mathrm{\mid }r\mathrm{\mid }\mathrm{\ne }0$, and $h:=s-1$. Then $s>1$ and $h>0$.

In case large, put $h^{\mathrm{\prime }}:=\mathrm{\mid }r\mathrm{\mid }-1$, so $h^{\mathrm{\prime }}>0$ and $\mathrm{\mid }r\mathrm{\mid }=1+h^{\mathrm{\prime }}$.

In case zero, for every rational $\epsilon >0$ and every $k\ge 1$ we have $\mathrm{\mid }{r}^k-0\mathrm{\mid }=\mathrm{\mid }0\mathrm{\mid }=0<\epsilon$, so $r^k\to 0$ and claim 1 holds.

In case small, $\mathrm{\mid }r{\mathrm{\mid }}^k{s}^k=(\mathrm{\mid }r\mathrm{\mid }s{)}^k=1^k=1$, so $\mathrm{\mid }r{\mathrm{\mid }}^k=1\mathrm{/}{s}^k$, and $s^k>0$.

In case small, Bernoulli applied to $h>0\ge -1$ gives $s^k=(1+h{)}^k\ge 1+kh>kh>0$ for every $k\ge 1$, using $1>0$ and $kh>0$.

In case large, Bernoulli applied to $h^{\mathrm{\prime }}>0\ge -1$ gives $\mathrm{\mid }r{\mathrm{\mid }}^k=(1+h^{\mathrm{\prime }}{)}^k\ge 1+k{h}^{\mathrm{\prime }}$ for every $k\in \mathbb{N}$.

In case large, let $M\in \mathbb{R}$ be arbitrary and use [L7] to fix a natural $n\ge 1$ with $M\mathrm{/}{h}^{\mathrm{\prime }}<n$; then $M\le n{h}^{\mathrm{\prime }}$, since multiplying $M\mathrm{/}{h}^{\mathrm{\prime }}\le n$ by $h^{\mathrm{\prime }}>0$ preserves the inequality.

In case small, let $\epsilon >0$ be rational; then $\epsilon h>0$, so [L7] supplies a natural $N\ge 1$ with $1\mathrm{/}N<\epsilon h$, whence $1\mathrm{/}(Nh)\le \epsilon$ on multiplying by $1\mathrm{/}h>0$.

In case small, combining steps 3.2 and 3.3: $0<kh<s^k$ gives $\mathrm{\mid }r{\mathrm{\mid }}^k=1\mathrm{/}{s}^k<1\mathrm{/}(kh)$ for every $k\ge 1$.

In case large, for every $k\ge n$ we have $k{h}^{\mathrm{\prime }}\ge n{h}^{\mathrm{\prime }}\ge M$, so $\mathrm{\mid }r{\mathrm{\mid }}^k\ge 1+k{h}^{\mathrm{\prime }}\ge 1+M>M$, the last step because $1>0$.

In case small, for every $k\ge N$ we have $kh\ge Nh>0$, hence $1\mathrm{/}(kh)\le 1\mathrm{/}(Nh)\le \epsilon$, and therefore $\mathrm{\mid }{r}^k-0\mathrm{\mid }=\mathrm{\mid }{r}^k\mathrm{\mid }=\mathrm{\mid }r{\mathrm{\mid }}^k<1\mathrm{/}(kh)\le \epsilon$.

In case large, an index $n$ has been produced for an arbitrary real $M$ with $\mathrm{\mid }r{\mathrm{\mid }}^k>M$ for all $k\ge n$, which is exactly divergence to $+\mathrm{\infty }$: claim 2 holds.

In case small, the rational $\epsilon >0$ was arbitrary and the index $N$ was produced from it, so $r^k\to 0$ and claim 1 holds.

The hypothesis $\mathrm{\mid }r\mathrm{\mid }<1$ of claim 1 is exhausted by cases zero and small, since $\mathrm{\mid }r\mathrm{\mid }\ge 0$ with $\mathrm{\mid }r\mathrm{\mid }=0$ exactly when $r=0$, so trichotomy leaves only $0<\mathrm{\mid }r\mathrm{\mid }<1$; the hypothesis $\mathrm{\mid }r\mathrm{\mid }>1$ of claim 2 is case large. Both claims are therefore established.