Beatty Log Series
Paul D. Hanna
pauldhanna at juno.com
Fri Jul 15 07:14:59 CEST 2005
Seqfans,
Can someone transform the following series:
Sum_{n>=1} ( 1/[n*x] - 1/(n*x) )
into one that converges more rapidly than
Sum_{n>=1} FractionalPart(n*x) / (n*x*[n*x])
for irrational x >1 ?
The series involves reciprocal terms of a Beatty sequence
and has the following interesting property.
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Given an irrational number t, 0<t<1,
define a function B(t) by the series
B(t) = Sum_{n>=1} ( 1/[n(1+t)] - 1/(n(1+t)) )
then
B(t) + B(1/t) = log(1+t) - log(t)*t/(1+t)
where [x] denotes the integer floor of x.
--------------------------------------------------
I do not know what B(t) is in itself,
but it can be expressed as:
B(t) = log(1+t)/(1+t) - L(t)
B(1/t) = log(1+1/t)/(1+1/t) - L(1/t)
where L is an unknown function that satisfies
L(1/t) = - L(t).
Now I wonder, what is the function L(t)?
Note: the sum of differences of reciprocal Beatty sequences
B(t) - B(1/t) = Sum_{n>=1} ( 1/[n(1+t)] - 1/[n(1+1/t)] )
= log(1+t)*(1-t)/(1+t) + log(t)*t/(1+t) - 2*L(t)
may be a clue to the mysterious L(t).
Perhaps there are 'constants' involved here
analogous to the Euler gamma constant ...
Taking the analogy even further,
wonder if a "Beatty Zeta function":
Z(s,t) = Sum_{n>=1} 1/[n*(1+t)]^s
where
Z(s,t) + Z(s,1/t) = zeta(s)
has been studied before (references?).
Any comments would be appreciated.
Thanks,
Paul
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