Show that $r^nrightarrow 0$ if $|r|

Question

Let $rin mathbb{C}$, $|r|<1$. Show that $r^nrightarrow 0$.
For that we have to show that $r^n$ is decreasing and bounded, right?
Let $a_n:=r^n$.
Then we have that $frac{a_{n+1}}{a_n}=frac{r^{n+1}}{r^n}=rleq |r|<1$ and so the sequence is decreasing.
Since $|r|<1$ we get that $-1<r<1Rightarrow (-1)^n<r^n<1^n Rightarrow (-1)^n<r^n<1$ and so the sequence is bounded.
So we know that the sequence converges to a limit $L$.
We have that $r^{n+1}=rcdot r^n$. Since $r^nrightarrow L$ we also have that $r^{n+1}rightarrow L$ and so we get for $nrightarrow infty$ that $L=rcdot LRightarrow Lcdot (r-1)=0$ and since $|r|<1$ and so $rneq 1$ it follows that $L=0$.
Is everything correct?

José Carlos Santos · Accepted Answer

No, it is not. Even assuming that $rinBbb R$, you did not prove correctly that the sequence is decreasing, and you could not have possibly have done so, since, if $-1<r<0$, the sequence is not decreasing (nor increasing, for that matter).
Besides, you don't have to prove that the sequence is bounded and decreasing.

Yves Daoust · Answer

As for any complex number $$|r^n-0|=||r|^n-0|$$ it suffices to show that the property holds in the reals.
Let $s:=|r|>0$. Now for any $epsilon>0$, $|s^n-0|<epsilon$ if $ngelog_sepsilon$, and the latter expression is always defined. (Needless to say, $s=0$ also works.)

math · Answer

So $|z^n - 0| = |z^n| = |z|^n rightarrow 0$, which is the definition of $z^n rightarrow 0$.

Show that $r^nrightarrow 0$ if $|r|<1$

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