Beatty Log Series - Correction

[Paul D. Hanna]

Correction: my statement 
 
> 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).
     
should have included -(1-t)/(1+t)/n  in the sum:
 
 B(t) - B(1/t) 
  = Sum_{n>=1} ( 1/[n(1+t)] - 1/[n(1+1/t)] - (1-t)/(1+t)/n )
  =  log(1+t)*(1-t)/(1+t) + log(t)*t/(1+t) - 2*L(t)
 
Can *this* sum be made to converge any faster?
 
Paul