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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!

shintuku
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    What do you mean by a sequence within an equation in that way? $\displaystyle \lim_{n \to \infty} \frac{a-ar^{n+1}}{1-r} = \frac{a}{1-r}$ means that $\displaystyle (\forall \varepsilon > 0)(\exists N \in \mathbb Z^+)(\forall n \in \mathbb Z^+) : n \geq N \implies \bigg| \frac{a-ar^{n+1}}{1-r} - \frac{a}{1-r} \bigg| < \varepsilon$. – azif00 Jun 24 '21 at 20:03
  • If your question really is "what exactly is the limit definition for $\lim_{n\to\infty} r^{n+1} = 0$ for $|r|<1$" then ask that. But of course try to find out whether it has already been asked here. – GEdgar Jun 24 '21 at 20:03
  • You will find many questions about the limit of a sequence on this site, including questions about its definition, for example https://math.stackexchange.com/questions/192586/can-you-help-me-understand-this-definition-for-the-limit-of-a-sequence. But I must point out that you will also find the definition of limit of a sequence in most any calculus textbook. – Lee Mosher Jun 24 '21 at 20:16
  • Also, you need the definition of limit of a sequence even just to formulate the definition of the sum of an infinite series. For example in your question we have $$\sum^\infty_{k=0} ar^k = \lim_{n \to \infty} \left(\sum^n_{k=0} ar^k\right) = \lim_{n \to \infty} \left(\frac{a-ar^{n+1}}{1-r}\right)$$ – Lee Mosher Jun 24 '21 at 20:18
  • Thank you all for the answers, it helped me get a better sense of what I was missing! – shintuku Jun 24 '21 at 20:44

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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.

Anik Bhowmick
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