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Hard Reciprocal Sum (Posted on 2020-03-13)

Evaluate the infinite sum

1 - 1/2 + 1/4 - 1/5 + 1/7 - 1/8 + ... + 1/(3n-2) - 1/(3n-1) + ...

See The Solution Submitted by Brian Smith

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Possible solution

| Comment 1 of 8

Both of the parts are infinite, but they grow very slowly, so we can take, say:

sum_(n=1)^100000 1/(3n-2)≈4.88165252611

sum_(n=1)^100000 1/(3n-1)≈4.27705384914

The difference = 0.60459867697. We cannot be sure about all the digits, so say 0.604599≈pi/(3*3^(1/2) by lookup.

WolframAlpha confirms that sum 1 to infinity (1/(3n-2)-1/(3n-1)) =π/(3sqrt(3))

Posted by broll on 2020-03-24 00:10:36

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