The Sum to Infinity

An endless sum with a total

Some sums go on forever. 1/2 + 1/4 + 1/8 + 1/16 + … never stops, because after every term there is another. Even so, its total does not grow without limit.

Think of a bar of length 1. The first term, 1/2, fills half of it. The next term, 1/4, fills half of what is left. The next, 1/8, fills half of what is left after that. The running totals are 1/2, 3/4, 7/8, 15/16, …, and after n terms the gap left is 1/2ⁿ. The gap halves at every step, so the totals get as close to 1 as you like, but they never pass it.

The series converges, and 1 is its sum to infinity.

After one, two and three terms, 1/2, then 3/4, then 7/8 of the bar is filled. Each new term fills half of the gap that is left.

When the total settles

Not every geometric series settles. With ratio 2, the series 1 + 2 + 4 + 8 + … adds bigger and bigger terms, and its running totals 1, 3, 7, 15, … pass every number you can name. That series diverges: it has no sum to infinity.

For the total to settle, the terms must shrink toward 0. In a geometric series they do exactly when the common ratio r lies strictly between −1 and 1, which is written |r| < 1. At r = 1 every term is the same, and the totals climb by the same step forever. At r = −1 the terms are 1, −1, 1, −1, …, and the totals jump between 1 and 0 without settling.

$S_{N} = \frac{1 - r ^{N}}{1 - r}, ∣ r ∣ \geq 1 : r^{N} \neq \to 0$

r = 1: every term is 1, Sₙ = N, and the sums climb a fixed step forever

Set r = 0.5 and take at least 10 terms

Each bar is a running total of 1 + r + r² + …. At r = 1 every term is 1, so the totals climb by 1 forever. Set r = 0.5 and take at least 10 terms: the totals climb toward 2 and never cross it.

The formula

The sum of the first n terms, from The Sum of a Geometric Series, is a(1 − rⁿ)/(1 − r). When |r| < 1, rⁿ gets closer and closer to 0 as n grows: halving again and again leaves almost nothing. So 1 − rⁿ gets closer and closer to 1, and the sum closes in on a/(1 − r).

This number is the sum to infinity, written S∞, so S∞ = a/(1 − r). It is valid only when |r| < 1. Outside that range the formula still produces a number, but the series has no sum, so the number means nothing. For 1 + 2 + 4 + …, the formula would give 1/(1 − 2) = −1, which cannot be the total of positive terms.

An example

In 8 + 4 + 2 + 1 + …, the first term is a = 8 and the common ratio is r = 4 / 8 = 1/2. The ratio lies between −1 and 1, so the sum to infinity exists: S∞ = 8/(1 − 1/2) = 8 ÷ 1/2 = 16. Dividing by 1/2 is the same as multiplying by 2, so the sum to infinity is twice the first term.

Check with the running totals: 8, 12, 14, 15, 15.5, … Each total is halfway from the one before to 16, and none of them passes 16.

The running totals of 8 + 4 + 2 + 1 + … on a number line: 8, 12, 14, 15. Each jump is half the one before, and the totals close in on 16.

Medicine in the body

In the next problem, the medicine in the body just after each dose is a geometric series. Its common ratio is the fraction of the medicine that remains from one day to the next.

Worked example: A Daily Dose of Medicine, and the Amount in the Body in the Long Run

Question A patient takes a 200 mg dose of a medicine once a day. By the time of the next dose, 20% of the medicine in the body remains and the rest has been cleared. (a) Find the amount of medicine in the body just after a dose in the long run. (b) After which dose is the amount just after the dose first within 1 mg of that long-run amount?

1.Just after the nth dose, the body holds the new 200 mg, plus 20% of the dose before it, plus 20% of 20% of the one before that, and so on back to the first dose. In mg this is 200 + 200(0.2) + 200(0.2)² + … + 200(0.2)ⁿ⁻¹, a geometric series with a = 200 and r = 0.2.

Just after a dose: the new 200 mg, plus 20% of the dose before, and so on.
2.Its sum is S_n = 200(1 − 0.2ⁿ)1 − 0.2 = 250(1 − 0.2ⁿ). For example, S₁ = 200, S₂ = 240 and S₃ = 248.

The sum is S_n = 250(1 − 0.2ⁿ): 200, 240, 248, … mg.
3.The common ratio is 0.2, and |0.2| < 1, so 0.2ⁿ gets closer and closer to 0 and the series has a sum to infinity: S_∞ = a1 − r = 2000.8 = 250.

Since |r| = 0.2 < 1, the sum to infinity is 2001 − 0.2 = 250.
4.(a) In the long run, the body holds 250 mg just after each dose.

(a) In the long run the body holds 250 mg just after each dose.
5.The gap between S_n and 250 mg is 250 − 250(1 − 0.2ⁿ) = 250 × 0.2ⁿ. After the 3rd dose it is 250 × 0.008 = 2 mg, and after the 4th dose it is 250 × 0.0016 = 0.4 mg.

The gap to 250 mg is 250 × 0.2ⁿ: 2 mg after dose 3, 0.4 mg after dose 4.
6.(b) The 4th dose is the first after which the amount, 249.6 mg, is within 1 mg of 250 mg.

(b) After the 4th dose the amount is 249.6 mg, the first within 1 mg of 250 mg.

Answer: (a) 250 mg; (b) after the 4th dose, when the amount is 249.6 mg

Common mistakes

Adding the doses as if none were cleared, which gives 200n mg and grows without limit. Only 20% of each dose is left a day later, so the older doses add less and less.
Using the fraction cleared, 0.8, as the common ratio, which gives 2001 − 0.8 = 1000 mg. The common ratio is the fraction that remains from one day to the next, which is 0.2.

More sequences and series problems, worked step by step →