Toolkit 4

Geometric series

1+r+r^2+\cdots+r^n=\frac{r^{n+1}-1}{r-1}

|r|<1:\quad 1+r+r^2+\cdots=\frac{1}{1-r}

Proof

Let

S = 1 + r + r^2 + \cdots + r^n.

Multiplying by $r$ , we get

rS = r + r^2 + r^3 + \cdots + r^{n+1}.

Subtracting the first equation from the second gives

(r-1)S = r^{n+1} - 1.

Therefore, for $r \ne 1$ ,

S = \frac{r^{n+1} - 1}{r - 1}.

If $|r| < 1$ , then $r^{n+1} \to 0$ as $n \to \infty$ . Hence,

1 + r + r^2 + \cdots = \lim_{n \to \infty} \frac{1 - r^{n+1}}{1 - r} = \frac{1}{1 - r}. \quad\square