# infinite summation of a function with exponential, power and a factorial.

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?

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 $\cdots ???$

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

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