I ran into this limit while evaluating an integral. <munder> <mo movablelimits="true" form="

Joel French 2022-07-03 Answered
I ran into this limit while evaluating an integral.
lim N ( 2 N + 1 n = 1 N 1 n )
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Answers (1)

jugf5
Answered 2022-07-04 Author has 18 answers
First of all we see that eta-function is defined (the series converges) for s > 0 (Dirichlet convergence test). Next, we use the relation between eta-function and zeta-function for s > 1
η ( s ) = n = 1 ( 1 ) n 1 n s = n = 1 1 n s 2 n = 1 1 ( 2 n ) s = ( 1 2 1 s ) ζ ( s )
Using this relation we can define zeta-function for s ( 0 ; 1 ) as ζ ( s ) = η ( s ) 1 2 1 s We can also present eta-function in the form
lim N n = 1 2 N ( 1 ) n 1 n s = lim N ( n = 1 2 N 1 n s 2 n = 1 N 1 ( 2 n ) s ) = lim N ( n = 1 N 1 2 1 s n s + n = N + 1 2 N 1 n s )
The second term can be presented as the Riemann sum at N
I 2 ( N ) = n = N + 1 2 N 1 n s = N N S 1 N n = N + 1 2 N 1 ( n N ) s N 1 s 1 2 d x x s = N 1 s 2 1 s 1 1 s
η ( s ) = lim N ( n = 1 N 1 2 1 s n s N 1 s 1 2 1 s 1 s )
ζ ( s ) = lim N ( n = 1 N 1 n s N 1 s 1 s ) ; s ( 0 ; 1 )
Our initial limit
lim N ( 2 N + 1 n = 1 N 1 n ) = lim N ( 2 N n = 1 N 1 n ) = ζ ( 1 2 ) = 1.46035...
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