1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

Hint

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 $\cdots ???$

$\endgroup$
0
$\begingroup$

Hint:

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 $~~b = \mu/(kT)$

$\endgroup$
4
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.