Geometric Series: Filling the Square

Fill a unit square by always taking half of what remains: the gap halves at every step and the total becomes 1.

12+14+18+⋯=1\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots=1

Examples & Exercises

Prove SN=1+r+⋯+rN=1−rN+11−rS_N = 1+r+\cdots+r^N = \frac{1-r^{N+1}}{1-r} (r≠1r \ne 1), and convergence to 11−r\frac{1}{1-r} for ∣r∣<1|r| < 1.

SN−rSN=1−rN+1S_N - rS_N = 1 - r^{N+1}
r=12r = \frac{1}{2}: only the gap (1/2)12(1/2)^{12} remains.
Show proof
  1. SN−rSN=(1+r+⋯+rN)−(r+⋯+rN+1)=1−rN+1S_N - rS_N = (1+r+\cdots+r^N) - (r+\cdots+r^{N+1}) = 1-r^{N+1}. Everything in the middle cancels.
  2. If r≠1r \ne 1, SN=1−rN+11−rS_N = \frac{1-r^{N+1}}{1-r}.
  3. If ∣r∣<1|r| < 1, rN+1→0r^{N+1} \to 0, so SN→11−rS_N \to \frac{1}{1-r}.

Prove 0.999…=10.999\ldots = 1 via the geometric series.

0.999⋯=9∑k=1∞(110)k=9⋅1/101−1/10=10.999\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=110r = \frac{1}{10}.
Show proof
  1. 0.999⋯=910+9100+⋯=9∑k=1∞(110)k0.999\cdots = \frac{9}{10}+\frac{9}{100}+\cdots = 9\sum_{k=1}^{\infty}\left(\frac{1}{10}\right)^k.
  2. By the previous formula with r=110r = \frac{1}{10}: the sum =9⋅1/101−1/10=9⋅19=1= 9\cdot\frac{1/10}{1-1/10} = 9\cdot\frac{1}{9} = 1.

Prove ∑n=1∞n2n=2\sum_{n=1}^{\infty}\frac{n}{2^n} = 2 with the S−S2S - \frac{S}{2} trick.

S−S2=∑n=1∞12n=1S - \frac{S}{2} = \sum_{n=1}^{\infty}\frac{1}{2^n} = 1
The same idea works with weights attached.
Show proof
  1. Put S=∑n=1∞n2nS = \sum_{n=1}^{\infty}\frac{n}{2^n}. Then S2=∑n=1∞n2n+1=∑n=2∞n−12n\frac{S}{2} = \sum_{n=1}^{\infty}\frac{n}{2^{n+1}} = \sum_{n=2}^{\infty}\frac{n-1}{2^n} (reindexing).
  2. Absolute convergence justifies termwise subtraction: S−S2=12+∑n=2∞12n=∑n=1∞12n=1S-\frac{S}{2} = \frac{1}{2}+\sum_{n=2}^{\infty}\frac{1}{2^n} = \sum_{n=1}^{\infty}\frac{1}{2^n} = 1.
  3. Hence S2=1\frac{S}{2} = 1, i.e. S=2S = 2.

6. Series