Zeta Function and Harmonic Series

Recently someone asked about ways of computing Euler's constant,
gamma.  This started me thinking about the partial sums of the
harmonic series.  Let h(n) denote the sum

            h(n)  =  1/1 + 1/2 + 1/3 + 1/4 + ... + 1/n

These sums are related to the zeta function by the formula 

                                         n   zeta(k)
      lim           h(2n) - h(n)  =    SUM  ---------
      n->oo                             k=2    2^k
                                      
Since zeta(k) approaches 1 as k increases, we can get faster 
convergence by re-writing this in the form

                                     1        n   zeta(k) - 1
     lim           h(2n) - h(n)  =  ---  +  SUM  -------------
     n->oo                           2       k=2      2^k

It seems that it should be possible to compute gamma by making use of 
this relation.  This could also be written as a recurrence

                                         2n    zeta(k) - 1
      lim    h(4n) = 2 h(2n) - h(n) +  SUM    -------------
      n->oo                             k=n+1      2^k

Notice that, for example, zeta(40) = 1.00000000000090949478..., so the
terms  (zeta(k)-1)/2^k  quickly become very small as k increases.