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Difficult sums

Hello. I'm interested in computing the sum \(\large \displaystyle \sum_{n=1}^\infty 2^{-2^n} \). Does anyone have any exact methods?

Note by Sal Gard
1 year, 6 months ago

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Your number is equal to the Fredholm number minus \(2^{-2^0} = \dfrac12\). Since Fredholm number has been proven to be a transcendental numbers, then there is no way of finding closed form for this number.

Pi Han Goh - 1 year, 6 months ago

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