Expand a limit form containing sum to series?

Questions
,
1

Hi, my problem is as follows?

Define F(q,\lambda) as

F(q,\lambda):=\lim\limits_{z\rightarrow 0}\left (\sum_{i=1/2}^\infty \ln [1+z^{-1}q^{i}\exp({-i^{2} \lambda})][1+zq^{i}\exp({i^{2} \lambda})]-\ln z\right ) ,

where the summation index i are half integers and |q|<1.
Expand F(q,\lambda) first as series of \lambda,

F(q,\lambda)=F_1(q)+F_2(q)\lambda^2+...

Then expand F_2(q) in q

F_2(q)=a_0+a_1q+a_2q^2+...

What I want is to compute the coefficients a_0,a_1,a_2 using Mathematica, and higher term’s coefficients if posible.
I tried the following code(rather straightforward) and it failed(got stuck in calculation)

Series[SeriesCoefficient[
Limit[Sum[Log[(1 + z q^i Exp[\[Lambda] i^2]) (1 + z^(-1) q^i Exp[-\[Lambda] i^2])], {i, 1/2, Infinity}] - 
Log[z], z -> 0], 
{\[Lambda], 0, 2}], {q, 0, 2}]

It seems mathematica is struggling to first do the infinite sum in the limit which possibly doesn’t have a simpler expression at all.

Any help or comment would be appreciated.

Background(hopefully useful to solve the problem):F(q,\lambda) is the free energy of a type of fermion system, F(q,\lambda)=\ln Z(q,\lambda) where Z is the partition function.

2

That seems to be difficult.

Look at (you forgot the Log in your code above)

sum[n_] := 
 Sum[Log[(1 + z q^i Exp[\[Lambda] i^2]) (1 + 
       z^(-1) q^i Exp[-\[Lambda] i^2])], {i, 1/2, n}] // Expand

and (you may want to chose different values for q, lambda and z )

Plot[Evaluate[
  sum[m] - Log[z] /. q -> .5 /. \[Lambda] -> 5 /. z -> 2], {m, 1, 15}]

Now look at

ser = Series[sum[3], {\[Lambda], 0, 3}] // FullSimplify

Perhaps you get some ideas when considering

Coefficient [ser, \[Lambda]^1]
Coefficient [ser, \[Lambda]^2]
3

I plot sum[n] using different q, lambda and z, and I find that when n goes to infinity, it simply diverges. So F(z,q,lambda) is not well defined. But my problem now is solved (in completely different way). Thank you for your advice.