# Seeking closed form of $\sum_{j=0}^\infty (-1)^j \binom nj /(e^{-j m+x}-1)$

Does this sum have closed form:

$$\sum_{j=0}^\infty \frac{(-1)^j \binom nj}{e^{-j m+x}-1}$$ Where: $e$ is exponential function and $x \in \mathbb{R},m>0,m \in \mathbb{R},n>0,n \in \mathbb{Z}$

Maybe this sum can use integral to solve it? Thank you

Update.

I'm seeking a n-th derivative.Using this formula:

$$\frac{\partial ^nf(x)}{\partial x^n}=\underset{m\to 0}{\lim }\, m{}^{\wedge}(-n)*\underset{j=0}{\overset{\infty }{(\sum }}(-1){}^{\wedge}j*\text{Gamma}(n+1)*f(-j*m+x)*(j!*\text{Gamma}(n+1-j){}^{\wedge}(-1))$$

where: $$f(x)=\frac{1}{e^x-1}$$

that way i'm need sum have closed form.

• I see two angles of attack: 1) either realizing it as complex integral (as you imagine it) with an infinite number of poles (theorem of residues) or 2) Poisson summation formula. Not enough time this evening to develop any of these. – Jean Marie Jan 23 '17 at 23:00
• If $a=1$, the term with $j=0$ has zero in its denominator. More generally, if $e^{jm} = a$, that term will have zero in its denominator. Is this a problem? – marty cohen Jan 23 '17 at 23:04
• a is a function exp(x) ;-) – Mariusz Iwaniuk Jan 23 '17 at 23:16

$\binom{n}{j}$ is zero if $j>n$, hence the given sum equals, assuming $e^x=a< 1$, $$-\sum_{j=0}^{n}(-1)^j\binom{n}{j}\frac{1}{1-ae^{-jm}}=-\sum_{h\geq 0}a^h\sum_{j=0}^{n}(-1)^j\binom{n}{j}e^{-jhm}=-\sum_{h\geq 0}a^h(1-e^{-hm})^n.$$ The other case ($a>1$) is similar. If $a=1$ the original sum, as well as any rearrangement, is diverging.
• And what if $a=1$? – marty cohen Jan 23 '17 at 23:06
• If $a=1$, the original sum has a zero denominator at the $j=0$ term. This seems to mean that it goes to infinity as $a \to 1$. Does your regrouping have the same property? – marty cohen Jan 23 '17 at 23:18
• @martycohen: that is kind of trivial. If $a=1$ the term $(1-e^{-hm})^n$ converges to $1$ for large values of $h$. – Jack D'Aurizio Jan 23 '17 at 23:20