I ran across a cool series I have been trying to chip away at. <munderover> &#x2211;<!-- ∑ -

Craig Mendoza 2022-06-27 Answered
I ran across a cool series I have been trying to chip away at.
k = 1 ζ ( 2 k + 1 ) 1 k + 2 = γ 2 6 ln ( A ) + ln ( 2 ) + 7 6 0.0786
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Answers (1)

feaguelaBapzo
Answered 2022-06-28 Author has 9 answers
We have
n = 2 ( n 3 log ( 1 1 n 2 ) n 1 2 n ) = lim N n = 2 N ( n 3 log ( 1 1 n 2 ) n 1 2 n ) = lim N ( n = 2 N n 3 log ( 1 1 n 2 ) ( N 2 + N 2 2 ) ( log ( N ) 2 1 2 + γ 2 + O ( 1 N ) ) ) = lim N [ n = 2 N ( 2 n 3 log ( n ) n 3 log ( n + 1 ) n 3 log ( n 1 ) ) ( N 2 + N 2 2 ) ( log ( N ) 2 1 2 + γ 2 ) ]
In the sum on the last line, we may gather together the coefficients of each logarithm (terms at the boundary of the sum are a little funny), giving
lim N [ n = 2 N ( 6 n log ( n ) ) + log ( 2 ) + ( N 3 + 3 N 2 + 3 N + 1 ) log ( N ) N 3 log ( N + 1 ) ( N 2 + N 2 2 ) ( log ( N ) 2 1 2 + γ 2 ) ] = lim N [ n = 2 N ( 6 n log ( n ) ) + log ( 2 ) N 3 log ( 1 + 1 N ) + ( 3 N 2 + 3 N + 1 ) log ( N ) ( N 2 + N 2 2 ) ( log ( N ) 2 1 2 + γ 2 ) ] = lim N [ n = 2 N ( 6 n log ( n ) ) + log ( 2 ) N 3 ( 1 N 1 2 N 2 + 1 3 N 3 + O ( 1 N 4 ) ) + ( 3 N 2 + 3 N + 1 ) log ( N ) ( N 2 + N 2 2 ) ( log ( N ) 2 1 2 + γ 2 ) ] = lim N [ n = 2 N ( 6 n log ( n ) ) + ( 3 N 2 + 3 N + 1 2 ) log ( N ) + ( 3 2 N 2 + 7 6 γ 2 + log ( 2 ) ) ] = 6 log ( lim N ( n = 1 N n n N N 2 / 2 + N / 2 + 1 / 12 e N 2 / 4 ) ) + 7 6 γ 2 + log ( 2 ) = 6 log ( A ) + 7 6 γ 2 + log ( 2 )
Here, I am taking
A = lim N n = 1 N n n N N 2 / 2 + N / 2 + 1 / 12 e N 2 / 4
as the definition of the Glaisher-Kinkelin constant.

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