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Rapsinincke 2022-07-02 Answered
Is this function
k 1 1 ( 2 k 1 ) s , R e ( s ) > 1
is well known?
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Answers (1)

Dayana Zuniga
Answered 2022-07-03 Author has 16 answers
Formally,
ζ ( s ) = k = 1 + 1 k s = k = 1 + 1 ( 2 k ) s + k = 1 + 1 ( 2 k 1 ) s = 1 2 s k = 1 + 1 k s + k = 1 + 1 ( 2 k 1 ) s
whence
k = 1 + 1 ( 2 k 1 ) s = ζ ( s ) ( 1 1 2 s ) .

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