Adding up forever
Tonight's idea
What could \(1 + \tfrac12 + \tfrac14 + \tfrac18 + \cdots\) mean? You cannot add infinitely many numbers. The honest definition: add the first \(n\) terms, call that the partial sum \(S_n\), and if the partial sums home in on a single number \(S\), the series converges to \(S\). Otherwise it diverges and the sum is simply not a number.
The geometric series \(a + ar + ar^2 + \cdots\) is the one you can finish by hand. Multiply \(S_n = a + ar + \cdots + ar^{n-1}\) by \(r\) and subtract:
\(S_n - rS_n = a - ar^n\), so \(S_n = \dfrac{a(1 - r^n)}{1 - r}\).
If \(\lvert r \rvert \lt 1\) then \(r^n \to 0\) and \(S_n \to \dfrac{a}{1 - r}\). So \(1 + \tfrac12 + \tfrac14 + \cdots = 2\), and \(0.999\ldots = \dfrac{9/10}{1 - 1/10} = 1\). Not close to 1. Equal to 1.
Now the trap. In the harmonic series \(1 + \tfrac12 + \tfrac13 + \cdots\) the terms shrink to zero, yet it diverges. Group the terms:
\[ 1 + \tfrac12 + \underbrace{\tfrac13 + \tfrac14}_{\gt\, 2 \cdot \frac14 = \frac12} + \underbrace{\tfrac15 + \cdots + \tfrac18}_{\gt\, 4 \cdot \frac18 = \frac12} + \underbrace{\tfrac19 + \cdots + \tfrac1{16}}_{\gt\, 8 \cdot \frac1{16} = \frac12} + \cdots \]Every bracket beats \(\tfrac12\). You collect halves forever, so the partial sums pass every number. Change \(\tfrac1k\) to \(\tfrac1{k^2}\) and it converges: Euler showed the sum is \(\dfrac{\pi^2}{6}\).
Watch one: a repeating decimal
\(0.272727\ldots = \tfrac{27}{100} + \tfrac{27}{10000} + \cdots\) is geometric with \(a = \tfrac{27}{100}\), \(r = \tfrac{1}{100}\). Sum \(= \dfrac{27/100}{99/100} = \dfrac{27}{99} = \dfrac{3}{11}\). Every repeating decimal is a fraction for exactly this reason.
Your turn
- Find the sums: \(3 + \tfrac34 + \tfrac{3}{16} + \tfrac{3}{64} + \cdots\) and \(1 - \tfrac13 + \tfrac19 - \tfrac{1}{27} + \cdots\). (What is \(r\) in the second one?)
- Turn \(0.\overline{36}\) and \(0.1\overline{6}\) (the bar marks the block that repeats) into fractions using geometric series. For the second, the repeating part starts one digit late.
- A ball is dropped from 10 m and each bounce brings it back to 60% of its previous height. Find the total vertical distance it travels. Count the downs and the ups separately.
- The grouping argument shows \(S_{2^k} \gt 1 + \tfrac{k}{2}\) for the harmonic series once \(k \ge 2\). How many terms does that guarantee are enough to pass 10? The true figure is 12,367. Is the guarantee close to the truth or far above it?
Stretch: a sum you can finish
Find \(S = \tfrac12 + \tfrac24 + \tfrac38 + \tfrac{4}{16} + \cdots = \sum \tfrac{k}{2^k}\). Hint: write \(S\) and \(\tfrac{S}{2}\) one above the other, lined up so the denominators match, and subtract.
Answers
- \(a = 3\), \(r = \tfrac14\): \(\dfrac{3}{1 - 1/4} = 4\). Second: \(a = 1\), \(r = -\tfrac13\): \(\dfrac{1}{1 + 1/3} = \tfrac34\).
- \(0.\overline{36} = \dfrac{36/100}{1 - 1/100} = \dfrac{36}{99} = \dfrac{4}{11}\). \(0.1\overline{6} = \tfrac{1}{10} + \dfrac{6/100}{1 - 1/10} = \tfrac1{10} + \tfrac{1}{15} = \tfrac16\).
- Down: \(10 + 6 + 3.6 + \cdots = \dfrac{10}{0.4} = 25\). Up: \(6 + 3.6 + \cdots = \dfrac{6}{0.4} = 15\). Total 40 m.
- Need \(1 + \tfrac{k}{2} \ge 10\), so \(k = 18\): \(2^{18} = 262{,}144\) terms. About twenty times the true 12,367. The grouping bound is sure but crude.
Stretch. \(S - \tfrac{S}{2} = \tfrac12 + \left(\tfrac24 - \tfrac14\right) + \left(\tfrac38 - \tfrac28\right) + \cdots = \tfrac12 + \tfrac14 + \tfrac18 + \cdots = 1\), so \(\tfrac{S}{2} = 1\) and \(S = 2\).