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If you add and multiply infinitely, it keeps increasing, doesn't it? And yet we have this formula, if $0<r<1, S_∞ = \dfrac{a_1}{1 - r}$. I can apply this formula, but can't wrap my head around it.

Sergey Zolotarev
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    You're not adding infinitely many times (addition is only defined for two numbers; by recursion/induction you can then add finitely many numbers). You're adding finitely many times, and taking limits. So, it's probably a good idea for you to systematically study limits first (also the condition in the beginning is written incorrectly; it should be $-1<r<1$ not $0>r>1$ which makes no sense). – peek-a-boo Apr 01 '22 at 16:50
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    https://en.wikipedia.org/wiki/Zeno%27s_paradoxes#Dichotomy_paradox – TheSilverDoe Apr 01 '22 at 16:51
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    Can you wrap your head around the decimal representation of $\frac{1}{3}$, or do you find that (which is just a particular example of a geometric series) equally counterintuitive? – Joe Apr 01 '22 at 16:57
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    What does geometric have to do with this? It seems to me that your doubt is how any infinite series with positive terms can have a finite sum. – Hans Lundmark Apr 01 '22 at 17:37
  • Walking from Point A to Point B you first travel 1/2 the distance, then 1/4 the distance (i.e. half of the remaining distance), then 1/8 the distance (i.e. half of the distance remaining after you travel 1/2 the distance and 1/4 the distance), etc. So if the distance between Point A and Point B is $x,$ we see that $x = \frac{1}{2}x + \frac{1}{4}x + \frac{1}{8}x + \ldots$ – Dave L. Renfro Apr 01 '22 at 17:42

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The above image is a visual representation of the infinite geometric series $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... = \sum_{n = 1}^\infty \left(\frac{1}{2} \right)^n$$

Think about it like this: As $n$ gets larger and larger, you have a consistently smaller area of the bigger square left. So when $n$ approaches infinity, there is nothing in the bigger square left, and the sum of the parts $=$ the area of the square i.e. $1$.

It is also important to consider the significance of the condition $-1 < r < 1$ here. If you had $r = 2$ for example, you would get the following series: $$2 + 4 + 8 + 16 + ...$$ This would indeed fit with your idea of adding infinitely many terms, where the series grows progressively larger and larger.

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Alternative demonstration around the series $$T = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots .$$

Let $S_n$ denote $~\displaystyle \sum_{i = 0}^n \frac{1}{2^i} ~: ~n \in \Bbb{Z_{\geq 0}}.$

It is easy to show, by induction, that $~\displaystyle S_n = 2 - \frac{1}{2^n}.$

Therefore, $~\displaystyle \lim_{n \to \infty} S_n = 2.$

Further, by definition, $~\displaystyle T = \lim_{n \to \infty} S_n.$

user2661923
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