If you wished you could formulate Sum[f, {k, 1, s n} ] by knowing information about how series f converges (if it does).

It might be that an outer formula could be equal ...

s Sum[f, {k, 1 n} ]
Sum[f, {k, 1 n} ]s^2

It might be that "s times n" contribute near nothing if n is already a large value, and so on.

ok. you said it isn't quantum physics but it heavily uses PDE. i agree it isn't quantum but it is not matter of simple sums (involves some equations used in quantum physics)

(it is a japanese study (partly funded by usa) which verifies equations and results of other studies using a custom "fractal technique", not the best paper to learn adsorption or math from certainly)

you realize that quantum equations on an exact experiment (a perfect carbon sheet with perfect adsorbate dispersion) will answer adsorption both exactly and very simply (by electrical pairing), while other technique will be judging empirical outcomes of particular applications (estimating situational lab outcomes) (knowing to what experiments these formulas apply essential, but not stated exactly in the paper - instead it analyzes "correct outcomes" of other publications. many conjectures are made "may not be simply caused by the tendency to trap into small pits on the carbon surfaces" (the original papers likely never professed they did). conclusion ends with a question: it may become possible to estimate accurately the fractal dimension of the adsorbing surface by the use of the n[Infinity] parameter.) I would like to highlight that with all that this "re-analysis" of past papers has not yet released new data as to expectations of previously published outcomes and papers.

https://en.wikipedia.org/wiki/Partition_function_(statistical_mechanics)

Here is the paper. (Removed due to copyright ownership issues - Moderators)

My biggest problem is that I really don't get how to make sense of the approximation that links equation (8) with equation (9):

Best regards,

Alberto Silva

I don't know. But the paper is published (or is it not?), and therefore available in any library. Publishing it here is only convenient avoiding the need to go to a library. So - go ahead.

Let me first say I'm "bad at chemistry"

"Because of the self-similarity in raggedness at various resolutions, adsorption of a bulky organic molecule sequesters several neighboring sites from binding. Based on this model, a generalized equation was derived that encompasses the Langmuir and Freundlich equations"

Your link leads to a copyright protected material so the context of all the meanings of all the variables are not forthcoming.

It appears to be a base simple counting equation that (will be / is) modified to fit the case of a measurement of "imperfect carbon <> organic adsorption, aqueous" (living / non-living may be a question). it's very specific. but not nearly so specific as to get into (atomic problems, Schrodinger).

"Chemical potential and chemical activity" to me are highly suspect (meaning, it is not testing a specific carbon grid with specific organic molecules at specific energies, it is a wide generalization of result of experiment). I only mention that because, being bad at chemistry, it interests me to learn about it.

Why do you think you cannot substitute a value (for each / every L) ? (yielding the answer for a given L) ?

L probably must be limited because if it were infinite (n would be infinite) but you'd have Inf in your binomial which wouldn't be a result a chemist was looking for.

Also: the last step shows the equation without summation (suggesting, without my having tested it, that it is simplified to be calculated without need of summation, that it converged to the function seen).

From calculus - the binomial series has a power series expansion (and so may be simplified inside a sum).

If you want to know how the sum was simplified ask that specifically.

OK, but what exactly is your problem?

Sorry, it is by no means clear what your problem is:

In[70]:= Z[N] = 
 Binomial[L/m, Nt]*Product[Binomial[m, n], {n, 1, Nt}]*q^Nt
S[N, mu] = Sum[Z[Nt]*a^Nt, {Nt, 0, L}]

Out[70]= (1/BarnesG[2 + Nt])I^(
 Nt (3 + Nt)) m^Nt q^Nt BarnesG[1 - m]^(-1 + Nt)
  BarnesG[2 - m]^-Nt BarnesG[1 - m + Nt] Binomial[L/m, Nt]

Out[71]= \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(Nt = 0\), \(L\)]\(
\*SuperscriptBox[\(a\), \(Nt\)]\ Z[Nt]\)\)

and

n[z_] := z^2;
z = 3;
Table[a[j], {j, 1, n[z]}]

Do[
t[j] = Table[a[z], {z, 1, n[j]}],
{j, 1, 4}]
t[2]