Geometric Series: Filling the Square

A unit square gets filled by always taking half of what remains. The gap halves at every step and the total becomes 1.

Paint half, then paint half of the rest. The painted area creeps closer and closer to 1. However small the gap gets, one step always remains. In partial sums, that leftover step just keeps halving. Added forever, the pieces make exactly 1.

For example, eat half of a 🍰 cake, then half of what is left. The bites go one half, one quarter, one eighth, yet together they approach the whole cake.

Examples & Exercises

Prove $S_N = 1+r+\cdots+r^N = \frac{1-r^{N+1}}{1-r}$ ( $r \ne 1$ ), and convergence to $\frac{1}{1-r}$ for $|r| < 1$ .

S_N - rS_N = 1 - r^{N+1}

r = \frac{1}{2}

: only the gap

(1/2)^{12}

remains.

Show proof

$S_N - rS_N = (1+r+\cdots+r^N) - (r+\cdots+r^{N+1}) = 1-r^{N+1}$ . Everything in the middle cancels.
If $r \ne 1$ , $S_N = \frac{1-r^{N+1}}{1-r}$ .
If $|r| < 1$ , $r^{N+1} \to 0$ , so $S_N \to \frac{1}{1-r}$ .

Prove $0.999\ldots = 1$ via the geometric series.

0.999\cdots = 9\sum_{k=1}^{\infty}\left(\frac{1}{10}\right)^k = 9\cdot\frac{1/10}{1-1/10} = 1

The same algebra with

r = \frac{1}{10}

Show proof

$0.999\cdots = \frac{9}{10}+\frac{9}{100}+\cdots = 9\sum_{k=1}^{\infty}\left(\frac{1}{10}\right)^k$ .
From the previous formula $\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$ , subtract the $k = 0$ term $1$ to get $\sum_{k=1}^{\infty} r^k = \frac{1}{1-r} - 1 = \frac{r}{1-r}$ (equivalently, factor out $r$ ). With $r = \frac{1}{10}$ : the sum $= 9\cdot\frac{1/10}{1-1/10} = 9\cdot\frac{1}{9} = 1$ .

Compute

Prove $\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$ with the $S - \frac{S}{2}$ trick.

S - \frac{S}{2} = \sum_{n=1}^{\infty}\frac{1}{2^n} = 1

The same idea works with weights attached.

Show proof

First, absolute convergence: $\frac{n}{2^n} \le \left(\frac{3}{4}\right)^n \iff n \le \left(\frac{3}{2}\right)^n$ .
At $n = 4$ , $4 \le \left(\frac{3}{2}\right)^4 = \frac{81}{16} = 5.0625$ ✓, and if $\left(\frac{3}{2}\right)^n \ge n$ then $\left(\frac{3}{2}\right)^{n+1} = \frac{3}{2}\left(\frac{3}{2}\right)^n \ge \frac{3}{2}n = n + \frac{n}{2} \ge n+1$ ( $n \ge 2$ ), so by induction $n \le \left(\frac{3}{2}\right)^n$ for all $n \ge 4$ .
Hence $\frac{n}{2^n} \le \left(\frac{3}{4}\right)^n$ for $n \ge 4$ ; since $\sum\left(\frac{3}{4}\right)^n$ converges (geometric, ratio $\frac34 < 1$ ), the comparison test gives absolute convergence of $\sum\frac{n}{2^n}$ .
$\frac{n}{2^n} \le \left(\frac{3}{4}\right)^n \quad (n \ge 4) \;\Longrightarrow\; \sum_{n=1}^{\infty}\frac{n}{2^n} \text{ converges absolutely}$
Put $S = \sum_{n=1}^{\infty}\frac{n}{2^n}$ . Then $\frac{S}{2} = \sum_{n=1}^{\infty}\frac{n}{2^{n+1}} = \sum_{n=2}^{\infty}\frac{n-1}{2^n}$ (reindexing).
Absolute convergence justifies termwise subtraction: $S-\frac{S}{2} = \frac{1}{2}+\sum_{n=2}^{\infty}\frac{1}{2^n} = \sum_{n=1}^{\infty}\frac{1}{2^n} = 1$ .
Hence $\frac{S}{2} = 1$ , i.e. $S = 2$ .

Answer

$2$