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How would I prove that $\{r^n\}_{n\in\mathbb{N}^+}$ converges to $0$ if $0 < r < 1$? I've tried using $\epsilon-N$ but the only way I can isolate $n$ is by getting $n > \frac{ln|\epsilon|}{ln|r|} = log_{r}|\epsilon|$. Is there something I'm not getting here? Am I allowed to choose $N > log_{r}|\epsilon|$?

xhsbm
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  • That's funny, because $\ln(\epsilon)/\ln(r)$ is positive for $0 < \epsilon, r < 1$. I think you should plot the function $x \mapsto \log_r(x)$ for some $0 < r < 1$ and have a look at the graph. – Marktmeister Oct 14 '20 at 07:18
  • Ok but it's still negative eventually, isn't it? I guess my question is if it's possible to prove this using $\epsilon-N$. I'll edit the question. – xhsbm Oct 14 '20 at 07:28
  • I already know we can prove this using some other theorems, but can I do it using $\epsilon-N$? – xhsbm Oct 14 '20 at 07:30
  • It is negative if $\epsilon > 1$. Since $0 < r < 1$, one has $r^n < 1 < \epsilon$ for every $n \geq 1$, so the the argument really works if you take $n > \log_r(\epsilon)$. – Marktmeister Oct 14 '20 at 07:31
  • Yes, I think so. – Marktmeister Oct 14 '20 at 07:33
  • Wait how does it work? I thought it was for all $\epsilon \in \mathbb{Q}^+$. Letting $\epsilon > 1$ means $n$ will be greater than a negative? That's not how the definition works isn't it? – xhsbm Oct 14 '20 at 07:35
  • Well, if $n$ is greater than a negative and is a natural number, then $n \geq 1$. ;-) – Marktmeister Oct 14 '20 at 07:36
  • Zero is a natural number that's greater than a negative so wouldn't it be $n\geq0$? – xhsbm Oct 14 '20 at 07:39
  • It really depends on your definition of the natural numbers, but if $\epsilon > 1$, then sure, you can take $n \geq 0$, too. – Marktmeister Oct 14 '20 at 07:40
  • So it doesn't matter if I get $n$ greater than some negative, because all naturals are greater than a negative? As long as I choose $N$ properly? – xhsbm Oct 14 '20 at 07:43

2 Answers2

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Here I recall the proof in the Rudin's book

let $k$ be a natural number and $p$ be a positive real number. For $n>2k$,
$\ (1+p)^n> \ _nC_k\ p^k= \frac{n(n-1)\ \cdots\ (n-k+1)} {k!}\ p^k>\ \frac {n^k\ p^k} {2^k\ k!}$
Therefore $0<\ \frac {1} {(1+p)^n} < \frac {2^kk!} {p^k} n^{-k}$
since $n^{-k} \xrightarrow {} 0\ $, ${(\frac {1} {1+p}) }^n \xrightarrow {} 0$
As we set $p$ be any positive real, $r^n$ converges to $0$ when $0<r<1$

prime235711
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You did everything correctly. I'm expanding on my comments.

We have to show that for every $\epsilon > 0$, there is a $N \geq 0$ such that for every $n \geq N$, one has $|r^n - 0| = |r^n| < \epsilon$.

If $\epsilon \geq 1$, then $\log_r(\epsilon) \leq 0$. So choosing $N = 1 > \log_r(\epsilon)$ yields that for every $n \geq N = 1$, we have $|r^n| < 1 \leq \epsilon$, which makes sense. So if $\epsilon \geq 1$, there is really nothing to show, since the sequence takes values between $0$ and $1$ anyways (except for $n = 0$).

So the really interesting case is $0 < \epsilon < 1$. In this case, taking $N > \log_r(\epsilon)$ is a real condition on $N$, since here, $\log_r(\epsilon)$ is positive. But then it goes through:

Choose $N > \log_r(\epsilon)$. For every $n \geq N$, one then has $n > \log_r(\epsilon)$ and hence $|r^n| < |r^{\log_r(\epsilon)}| = \epsilon$, perfect!

Marktmeister
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