This is in the context of trying to prove to what a geometric series $\sum^\infty_{k=0} ar^k$ converges to when $|r| < 1$. The complete argument goes:
$$ar^0 + ar^1 + ... + ar^n - ar^1 - ar^2 - ... - ar^{n+1}= a-ar^{n+1}$$ $$=\sum^n_{k=0} ar^k - r \sum^n_{k=0} ar^k= (1-r)\sum^n_{k=0} ar^k$$
which implies $$\sum^n_{k=0} ar^k = \frac{a-ar^{n+1}}{1-r}$$
from which we get $$\sum^\infty_{k=0} ar^k = \frac{a}{1-r}-\frac{a}{1-r}* \lim \limits_{n \to \infty} r^{n+1} = \frac{a}{1-r}$$
My question is, what exactly is the limit definition for $\lim\limits_{n\to\infty}r^{n+1}=0$, where $|r| < 1$, that I need in order to do this last step? Since at no point we're talking about a function, it seems the epsilon-delta limit definition isn't relevant, and I was thinking perhaps it could be that of a sequence, but then, I have never used a sequence within an equation in that way, so I'm not too sure how to proceed. Does anyone have a hint?
Thank you for your time!
In order to find $\lim\limits_{n \to \infty} r^{n+1}$, we need to find $\lim\limits_{n \to \infty} r^{n}$ when $|r|<1$. Clearly when $r=0$, the sequence is $\{0, 0, 0, ... ...\}$, which converges to $0$.
Now we will deal with the situation when $0<|r|<1$. This gives $\frac{1}{|r|}>1$. Let $\frac{1}{|r|}=a+1$, where $a>0$. Then $|r^n - 0|=|r^n|=|r|^n=\frac{1}{(a+1)^n}>na \;\forall n \in \mathbb N$. So $|r^n-0|<\frac{1}{na}\;\forall n \in \mathbb N$.
Let $\epsilon > 0$. Then $|r^n - 0|<\epsilon$ holds if $n > \frac {1}{a \epsilon}$.
Let $k = [\frac{1}{a \epsilon}]+1$. Then $k$ is a natural number and $|r^n-0|< \epsilon\; \forall n \geq k$
Since $\epsilon$ is arbitrary, $\lim\limits_{n \to \infty} r^{n}=0$. Now combine two cases.