Every decimal expansion represents a series. For example: $0.333\ldots = \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots = \sum_{n=1}^\infty \frac{3}{10^n} = 3\sum_{n=1}^\infty \frac{1}{10^n}$

This is a geometric series with $a = \frac{1}{10}$ and $r = \frac{1}{10}$: $= 3 \cdot \frac{1/10}{1 - 1/10} = 3 \cdot \frac{1}{9} = \frac{1}{3}$

Thus $0.\overline{3} = \frac{1}{3}$.

If principal $P$ earns interest rate $r$ compounded $n$ times per year, the amount after $t$ years is: $A = P\left(1 + \frac{r}{n}\right)^{nt}$

As $n \to \infty$ (continuous compounding): $A = P\lim_{n \to \infty}\left(1 + \frac{r}{n}\right)^{nt} = Pe^{rt}$

This involves the fundamental limit $\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$.

Achilles runs 10 times as fast as a tortoise. The tortoise starts 100 meters ahead. When will Achilles catch the tortoise?

Achilles must cover: $100 + 10 + 1 + 0.1 + 0.01 + \cdots$ meters.

This is a geometric series: $\sum_{n=0}^\infty \frac{100}{10^n} = \frac{100}{1 - 1/10} = \frac{100 \cdot 10}{9} = \frac{1000}{9} \approx 111.1 \text{ meters}$

Achilles catches the tortoise after running approximately 111.1 meters.

A biased coin has probability $p$ of heads. What's the expected number of flips until the first heads?

Let $X$ = number of flips. Then: $E[X] = \sum_{n=1}^\infty n \cdot P(X = n) = \sum_{n=1}^\infty n(1-p)^{n-1}p$

Factor out $p$: $= p\sum_{n=1}^\infty n(1-p)^{n-1}$

This is the derivative of a geometric series. Let $S = \sum_{n=0}^\infty (1-p)^n = \frac{1}{1-(1-p)} = \frac{1}{p}$.

Differentiating: $\sum_{n=1}^\infty n(1-p)^{n-1} = \frac{d}{dp}\left[\frac{1}{p}\right] = -\frac{1}{p^2} \cdot (-1) = \frac{1}{p^2}$.

Wait, let me recalculate. Actually, using the formula for $\sum_{n=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}$: $E[X] = p \cdot \frac{1}{[1-(1-p)]^2} = p \cdot \frac{1}{p^2} = \frac{1}{p}$

For a fair coin ($p = 1/2$), the expected number of flips is $\frac{1}{1/2} = 2$.

Euler famously solved the Basel problem: what is $\sum_{n=1}^\infty \frac{1}{n^2}$?

Using Fourier series techniques, Euler showed: $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.6449$

This connects the integers to the circle constant $\pi$ in a surprising way.

The integral $\int_0^1 e^{-x^2}\,dx$ has no elementary antiderivative, but we can use series: $e^{-x^2} = \sum_{n=0}^\infty \frac{(-x^2)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}$

Integrating term-by-term: $\int_0^1 e^{-x^2}\,dx = \sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^1 x^{2n}\,dx = \sum_{n=0}^\infty \frac{(-1)^n}{n!(2n+1)}$

$= 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} - \cdots \approx 0.7468$