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By "Riemann Rearrangement Theorem", we can rearrange the conditionally convergence series and make it equal to arbitrary value.

And Riemann showed that rearranging the alternating harmonic series by $ {1 \over 2} \ln(2) $.

But I wondered what specific way to rearrange the certain series by arbitrary value.

such as:

$$ 1 - {1 \over 2} + {1 \over 3} - {1 \over 4} + \cdots \text{ → (Rearrange)} = N, $$ $$ \text{where } N \text{ is arbitrary real number} $$

Is there a specific way?

Thank you for reading.

Antithesis
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1 Answers1

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You have $$ 1 + \frac 1 3 + \frac 1 5 + \frac 1 7 + \cdots = +\infty $$ $$ -\frac 1 2 - \frac 1 4 - \frac 1 6 - \frac 1 8 - \cdots = -\infty $$ Suppose I want a rearrangement that makes the sum equal to $10.$

Keep adding up odd terms until the sum exceeds $10.$ Then add one even term so that the sum is less than $10.$ Then keep adding up odd terms until the sum exceeds $10.$ Then add one even term so that the sum is less than $10.$ Then keep adding up odd terms until the sum exceeds $10.$ Then add one even term so that the sum is less than $10.$ And so on.

This will converge to $10.$

Michael Hardy
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