The Sum of a Geometric Series

Multiply the sum by the ratio

In 1 + 2 + 4 + 8 the terms double, so pairing the ends does not work: 1 + 8 = 9, but 2 + 4 = 6. A geometric series needs a different idea.

Call the sum S, so S = 1 + 2 + 4 + 8. Multiply every term by the common ratio, 2: 2S = 2 + 4 + 8 + 16. Now write the two sums one above the other, with equal terms lined up. The terms 2, 4 and 8 are in both.

The sum S and the doubled sum 2S lined up, one power of 2 in each column. Doubling moves every term one column along, so 2, 4 and 8 are in both rows. Only the 1 and the 16 are not shared.

Subtract, and the middle cancels

Subtract S from 2S: (2 + 4 + 8 + 16) − (1 + 2 + 4 + 8). The 2, the 4 and the 8 are added in the first bracket and taken away in the second, so they cancel. Only 16 − 1 is left. On the left side, 2S − S = S, so S = 16 − 1 = 15.

Check by adding: 1 + 2 + 4 + 8 = 15. The sum is the term after the last one, 16, minus the first term, 1.

The same steps in letters

A geometric series with first term a, common ratio r and n terms is $S=a+a\times r+a\times r^2+\cdots +a\times r^{n-1}$S = a + a × r + a × r² + … + a × rⁿ⁻¹. Multiply every term by r: $r\times S=a\times r+a\times r^2+\cdots +a\times r^{n-1}+a\times r^n$r × S = a × r + a × r² + … + a × rⁿ⁻¹ + a × rⁿ. Every term from a × r to $a\times r^{n-1}$a × rⁿ⁻¹ is in both sums, so subtracting S from r × S leaves $r\times S-S=a\times r^n-a$r × S − S = a × rⁿ − a.

r lots of S, take away 1 lot of S, leaves r − 1 lots of S, so r × S − S = S(r − 1). In the same way, $a\times r^n-a=a\left(r^n-1\right)$a × rⁿ − a = a(rⁿ − 1). So $S\left(r-1\right)=a\left(r^n-1\right)$S(r − 1) = a(rⁿ − 1). Divide both sides by r − 1 to get the sum of n terms: $S=\frac{a\left(r^n-1\right)}{\left(r-1\right)}$S = a(rⁿ − 1)/(r − 1).

This needs r to be different from 1, because dividing by 0 is impossible. When r = 1 every term is a, and the sum is simply n × a.

Check the formula

For 3 + 6 + 12 + 24, a = 3, r = 2 and n = 4. The formula gives $3\times \frac{\left(2^4-1\right)}{\left(2-1\right)}=3\times \frac{15}{1}=45$3 × (2⁴ − 1)/(2 − 1) = 3 × 15/1 = 45, and adding directly gives 3 + 6 + 12 + 24 = 45.

The formula also works when the ratio is less than 1, but then $r^n-1$rⁿ − 1 and r − 1 are both negative. Multiplying the top and the bottom of the fraction by −1 gives the same formula in the form $S=\frac{a\left(1-r^n\right)}{\left(1-r\right)}$S = a(1 − rⁿ)/(1 − r), which keeps the numbers positive. For $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}$1/2 + 1/4 + 1/8 + 1/16, $a=\frac{1}{2}$a = 1/2, $r=\frac{1}{2}$r = 1/2 and n = 4, so the top of the fraction is $\frac{1}{2}\times \left(1-\frac{1}{16}\right)=\frac{1}{2}\times \frac{15}{16}=\frac{15}{32}$1/2 × (1 − 1/16) = 1/2 × 15/16 = 15/32, and the bottom is $1-\frac{1}{2}=\frac{1}{2}$1 − 1/2 = 1/2. Dividing $\frac{15}{32}$15/32 by $\frac{1}{2}$1/2 doubles it, so $S=\frac{15}{16}$S = 15/16.

After 4 steps the colored part is $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{15}{16}$1/2 + 1/4 + 1/8 + 1/16 = 15/16, and the gap is $\frac{1}{16}$1/16. Take 8 steps: the formula says the colored part is $1-\frac{1}{2^8}=\frac{255}{256}$1 − 1/2⁸ = 255/256.