difficult/unconventional series expansion

[Roman]

Dear group,

I am trying to do an unconventional series expansion, and find it
difficult with Mathematica. Given the function
S[x_] = Log[E^x-1]/(E^x-1)+(1+1/(E^x-1))Log[1+1/(E^x-1)]
I am looking for the behavior for very large x.
[For the interested: S(x) is the entropy (in units of the Boltzmann
constant) of a harmonic oscillator, with x=\hbar\omega/(kT). So I'm
looking for the low-temperature behavior of the entropy.]

The problem is that simply doing
Series[S[x],{x,Infinity,1}]
does nothing (since this is not a series expansion in the usual
sense). But I know what I would like to find: the lowest terms of the
"series expansion" are
(x+1)E^(-x+(2x+1)/(2(x+1))E^(-x))
Do you know how to find this expression automatically in Mathematica?
I am interested in a general technique, not just the results for this
particular function.

Cheers!
Roman.

[David Park]

What about:

S[x_] := Log[E^x - 1]/(E^x - 1) + (1 + 1/(E^x - 1)) Log[
1 + 1/(E^x - 1)]

S2[y_] = S[1/y];
Series[S2[y], {y, y0, 1}] // Normal;
Limit[%, y0 -> 0, Direction -> -1]

0

If you approach from the other direction you obtain an imaginary value.

David Park
djm...@comcast.net
http://home.comcast.net/~djmpark/

[Roman]

David,
it is clear that S[x] -> 0 as x->inf. What I am looking for is an
expression for the deviation of the function from this asymptotic
behavior, similar to the expression I suggest in the second paragraph.
Roman.