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August 13th, 2018, 01:08 AM   #1
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[2;3,5,7,11,13,17,...]=?

Does anyone know how to calculate the exact value of this continued fraction [2;3,5,7,11,13,17,...] ?
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August 13th, 2018, 06:03 AM   #2
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My guess is that it can't be simplified.
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August 22nd, 2018, 07:26 PM   #3
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Wild guess: log(exp(1)+1)+1
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August 22nd, 2018, 10:17 PM   #4
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You can using this formula
S=a/1-r
where S=sum of terms to infity
a=first term
r=common ratio
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August 23rd, 2018, 05:58 AM   #5
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There isn't a common ratio in this case.
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