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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.

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    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. – 2017-01-23
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    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? – 2017-01-23
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    a is a function exp(x) ;-) – 2017-01-23

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$\binom{n}{j}$ 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.

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    And what if $a=1$? – 2017-01-23
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    @martycohen: typo fixed and case included. – 2017-01-23
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    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? – 2017-01-23
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    @martycohen: that is kind of trivial. If $a=1$ the term $(1-e^{-hm})^n$ converges to $1$ for large values of $h$. – 2017-01-23
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    @Jack D'Aurizio .'a' is a function exp(x) ,sorry i'm update the question. – 2017-01-23
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    @MariuszIwaniuk: no need to update. It is enough to state $a=e^x$. – 2017-01-23
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    This seems like a possibly useful rearrangement, but is it a closed form? – 2017-01-24
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    @DavidK: that depends on your concept of *closed form*. It is not "more closed" than the original sum (that only has a finite number of non-zero terms), but it depends on a series with positive coefficients that is pretty easy to approximate. We cannot say the same for the original sum. – 2017-01-24
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    OK, instead of "possibly useful" I should have said "clearly useful." The answer does not always have to be exactly what the asker expected. – 2017-01-24
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    @Jack D'Aurizio. For me "not useful" answer :( – 2017-01-24
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    @MariuszIwaniuk: sorry, that is the best I can provide. It would be useful to know more exactly what you are looking for. – 2017-01-24
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    @MariuszIwaniuk: anyway, there is a way shorter approach for computing the $n$-th derivative of $\frac{1}{e^x-1}$. Expand it as a geometric series, differentiate it termwise, exploit Faulhaber's formula. Bernoulli numbers are obviously involved. – 2017-01-24
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    Your comment was very helpful. So you have a 1+ from me.:) – 2017-01-25