How to understand the counterintuitive fact that you can calculate an infinite geometric series?

Question

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.

Answer 1

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.

Answer 2

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