It seems I managed to solve the problem in the same way as previously, that is by substituting z=1/s and calculating the coefficients of the ordinary Taylor series. But the necessary trick is to first divide the function by s^(3/2), to eliminate the singularity. By using the following code (version 2) I obtained 50 expansion coefficients in about 25 minutes. I still do not understand why Series[] is so much slower. Leslaw. The code:

nmax=20;
fun[s_]:=(((1+s)^2+3*s^(3/2)*(1+s)^(1/2)*th-4*s*(1+s)*th+3*s^(1/2)*(1+s)^(3/2)*th+s^2*th^2))/(8 (s^(3/2)*(1+s)^(1/2)*((1+s)^(1/2)+s^(1/2)*th)^3));
(* Version 1: calculation of the expansion coefficients using Series[] --------  *)
coeffs1=Table[Simplify[SeriesCoefficient[Series[fun[s],{s,∞,nmax}],n+1/2]],{n,0,nmax}]
(* Version 2: manual calculation of the expansion coefficients ----------------  *)
terms=Array[F,nmax+1,0];
(* Note the trick: I multiply fun by s^(3/2) because the leading expansion term is ~s^(-3/2)  *)
F[0]=Simplify[ReplaceRepeated[fun[s]*s^(3/2),{s->1/z}],Assumptions->{z>0}];
Do[F[k]=Simplify[D[F[k-1],z]],{k,1,nmax}];
Do[F[k]=(F[k]/(k!)),{k,0,nmax}];
coeffs2=Simplify[ReplaceRepeated[terms,{z->0}]]

You have not copied the entire expression for F[s] (only the denominator). The expression is more complicated. For the entire expression and 40 terms I waited for about half an hour and aborted the calculations. Some nine months ago I had a similar problem with a much simpler formula, and then I had to wait in vain for three days (in a batch mode) as far as I remember. A replacement of the Series[] command by a recursive calculation of successive derivatives (after converting the problem to a Taylor expansion into powers of 1/s) then reduced the timing to a minute or so. I don't know how Series[] works (I'd love to, by the way, but this seems to be a secret), but in my opinion it either does a lot of unnecessary things, or is horribly unoptimal. Leslaw