I solved an assignment problem in statistical thermodynamics where I need to find out the following summation to evaluate a property.

$$\sum_{n = 1}^{+\infty} \left(\frac{L}{\lambda}\right)^n \frac{1}{n!} \large e^{\frac{\mu n}{kT}}$$

In the equation, all parameters like L, k, T etc are constants. Is there any general expansion formula to simplify this summation?



Using simpler notations, you are looking for $$\sum_{n=1}^\infty \frac{a^n}{n!}e^{bn}$$ Write $$a^n e^{bn}=e^{(b+\log(a))n}$$ Now $\cdots ???$



Use C. Leibovici's hint and your knowledge, in order to get the final answer for the general sum, which is:

$$\sum_{n = 1}^{+\infty} \frac{a^n}{n!} e^{bn} = e^{a e^b}-1$$

Where in your case $a = L/\lambda~~$ and $~~b = \mu/(kT)$

  • 1
    $\begingroup$ He doesn't has $e^{bn}$ with $b\neq 0$, does he? $\endgroup$ – SK19 Feb 21 '18 at 9:57
  • $\begingroup$ "Doesn't has"? - Anyway I fixed it, I forgot $n$ whilst editing. $\endgroup$ – Turing Feb 21 '18 at 9:59
  • $\begingroup$ Geez, non native languages in the morning, right? Anyway, that explains it :) $\endgroup$ – SK19 Feb 21 '18 at 11:34
  • $\begingroup$ Does this form of expansion have some nomenclature...for example Euler's summation? $\endgroup$ – Sriram Krishnamurthy Feb 21 '18 at 12:00

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