perplexus.info :: Calculus : Hard Reciprocal Sum

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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Solution

| Comment 2 of 8 |

Series = Sum 1/( 9(n-1/3)(n-2/3) ) = pi sqr(3)/9

Posted by FrankM on 2020-04-05 12:49:47

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