Hey I've been having some trouble figuring out this problem. Not sure what to do with the restriction on $r$, or really how to go about writing it formally. Our professor assigned this as a problem but its not even mentioned Epsilon Delta Proofs aren't even mentioned in our textbook.
Give an $\epsilon-\delta$ proof that $$\lim_{n\to \infty} r^n = 0$$ when $|r|<1$ and $r^n$ is divergent when $|r|>1$.
Thank you so much for any help.
On
Repeating a proof I saw in "What is Mathematics":
This proof is for $0 < r < 1$; for negative $r$, use $|r^n| = |(-r)^n|$.
Since $0 < r < 1$, $r = \frac1{1+a}$ where $a =\frac1{r}-1 > 0 $.
By Bernoulli's inequality (one form of which is $(1+a)^n \ge 1+an$ for $a \ge 0$ and integer $n \ge 1$), for $n \ge 1$, $(1+a)^n \ge 1+an >an $ so $r^n =\frac1{(1+a)^n} < \frac1{an} = \frac1{\left(\frac1{r}-1\right)n} $.
This should make completing your $\delta-\epsilon$ proof easier, and it doesn't need logs.
Proof of Bernoulli's inequality:
Since $1+a \ge 1+a$, true for $n = 1$.
If true for $n$, $(1+a)^{n+1} =(1+a)^{n}(1+a) \ge (1+na)(1+a) =1+(n+1)a +a^2 > 1+(n+1)a $.
Saying that $\lim_{n\to\infty}r^n=0$ (for $|r|<1$) is an abbreviation for a longer statement:
There is no problem if $r=0$, so we can assume $r\ne0$. Since $|r|<1$, we have that $|r|\cdot|r|<|r|\cdot 1$, or $|r|^2<|r|$. From this we can continue and conclude (by induction) that $|r|^{n+1}<|r|^n$ for every positive integer $n$. This is the same as saying $|r^{n+1}|<|r^n|$. So, in order that the inequality above is satisfied, we just need to find $m$ such that $|r^m|<\varepsilon$
This is the same as requiring that $m\log|r|<\log\varepsilon$ or, since $|r|<1$ so that $\log|r|<0$, $$ m>\frac{\log\varepsilon}{\log|r|} $$ Such an integer surely exists. (The logarithm is in basis $e$ or any basis $>1$.)
Saying that $r^n$ diverges for $|r|>1$ means that you can find no number $l$ such that $\lim_{n\to\infty}r^n=l$ (with the definition above).
Hint: for $|r|>1$, we have $|r|^{n+1}>|r|^n$ for every positive integer $n$.