MISSION · Calculus
When an Infinite Sum Has an Answer
Adding forever and landing on a number
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The idea in one sentence
An infinite sum means the limit of its partial sums — and that limit sometimes exists, which is stranger than it sounds.
Partial sums are the definition
You cannot add infinitely many numbers one at a time. So the sum is defined as what the running total approaches:
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots\]The running totals are \(0.5,\; 0.75,\; 0.875,\; 0.9375, \ldots\) — each step closes half the remaining gap to 1, and never overshoots. The limit is exactly 1, so the series converges to 1.
This is Zeno’s paradox with the arithmetic done. Crossing a room by halves takes infinitely many steps and a finite amount of time, because the steps shrink fast enough for the total to be finite.
Geometric series: a complete answer
For \(a + ar + ar^2 + \cdots\) the question is fully settled by the ratio:
\[\text{converges if } |r| < 1, \quad \text{to } S = \frac{a}{1 - r}\]| If $$ | r | \ge 1$$ the terms don’t shrink towards zero and the total runs away. |
The formula turns recurring decimals into fractions. \(0.3333\ldots\) is \(0.3 + 0.03 + 0.003 + \cdots\) with \(a = 0.3\) and \(r = 0.1\), so the sum is \(\frac{0.3}{0.9} = \frac{1}{3}\). It also proves \(0.999\ldots = 1\) exactly, not approximately — \(a = 0.9\), \(r = 0.1\), sum \(\frac{0.9}{0.9} = 1\).
Shrinking terms is not enough
Here’s the result that overturns most people’s intuition.
\[1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\]The terms shrink to zero, so surely the total settles down? It doesn’t. This harmonic series diverges — the partial sums grow without any bound at all, just extremely slowly. Reaching a total of 100 takes roughly \(10^{43}\) terms, but it gets there.
The classic proof groups the terms: \(\frac{1}{3} + \frac{1}{4}\) exceeds \(\frac{1}{2}\), the next four terms also exceed \(\frac{1}{2}\), the next eight likewise, and you can do that forever. Infinitely many halves cannot converge.
Compare that with \(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots\), which does converge — to \(\frac{\pi^2}{6}\), an answer nobody expected the first time it was found. Squaring the denominators is just enough extra shrinkage to make the difference.
The test that only rules things out
If the terms don’t approach zero, the series diverges. That’s reliable, and it’s the first thing to check.
The converse is false, as the harmonic series shows. Terms going to zero is necessary but not sufficient, and treating it as sufficient is the standard mistake in this topic.
Common mix-ups
- Assuming shrinking terms guarantee convergence. The harmonic series says no.
- Using the geometric formula when \(|r| \ge 1\), which produces a confident wrong number.
- Confusing the sequence with the series. The sequence \(\frac{1}{n}\) converges to 0; the series \(\sum\frac{1}{n}\) diverges.
- Treating \(0.999\ldots\) as slightly less than 1. It is 1.
Check yourself
Does \(8 + 4 + 2 + 1 + \cdots\) converge? If so, to what?
The ratio is \(\frac{1}{2}\), so yes. \(S = \frac{8}{1 - 0.5} = 16\).
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