Weird fractions

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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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#1 h Mar 30, 2025, 3:15 AM

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While I was doing a question I made this really weird observation:

So first, we suppose $S$ is the infinite sum $\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...$ . Then $S$ is more than $0$ since $\frac{1}{1}>\frac{1}{2}$ , $\frac{1}{3}>\frac{1}{4}$ , and so on. But we can rewrite it as $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...-(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...)=0.$ So is $S$ more than $0$ or equal to $0$ ? Help is much appreciated

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Facejo

#2 h Mar 30, 2025, 3:27 AM • 1 Y

Y by wangyanliluke

Your series is a conditionally converging series (in fact, it's the alternating harmonic series, the "canonical" example), so, by the Riemann Rearrangement Theorem, you can literally get the series to equal any real value by shuffling around the terms. (That is to say, if the series is conditionally converging, you can't arbitrarily exchange terms in order to find the sum.)

This post has been edited 1 time. Last edited by Facejo, Mar 30, 2025, 3:27 AM

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