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Is there a closed form of the following sum? $$\sum^{m}_{j=0}\frac{(-1)^{j}{m \choose j}}{n+jk}$$ I figure it should but the binomial is throwing me off. Any help would be greatly appreciated.

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    $\begingroup$ Uh, so is $k$ just a constant here? Or were you $jk$ about there being a $k$? $\endgroup$ – Chris Jul 4 '17 at 20:11
  • $\begingroup$ Sorry for not clarifying, n and k are both constants in this sum. I appreciate the pun. $\endgroup$ – JRose Jul 4 '17 at 20:14
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    $\begingroup$ I guess what I'm really wondering, OP, is whether this is a component of a larger sum or expression that people might be more successful trying to help you simplify. (Otherwise the appearance of $n$ and $k$ would be a little odd.) $\endgroup$ – Chris Jul 4 '17 at 20:18
  • $\begingroup$ It is a component of a larger sum, I'll add it in so the variables make more sense in context. $\endgroup$ – JRose Jul 4 '17 at 20:38
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The given sum equals $$ \int_{0}^{1}\sum_{j=0}^{m}\binom{m}{j}(-1)^j x^{jk+n-1}\,dx =\int_{0}^{1}x^{n-1}(1-x^k)^m\,dx$$ and by the substitution $x=z^{1/k}$ and Euler's Beta function this equals $\frac{\Gamma(m+1)\,\Gamma\left(\frac{n}{k}\right)}{k\,\Gamma\left(1+m+\frac{n}{k}\right)}.$

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Assuming $n, k, m$ are positive integers, you could write this as

$$\log \left(\dfrac{\prod_{j \text{ even}} (n+jk)^{{m \choose j}}}{\prod_{j \text{ odd}} (n+jk)^{{m \choose j}}}\right)$$

For example, if $m=4$ it is $$\log \left( {\frac {n \left( 2\,k+n \right) ^{6} \left( 4\,k+n \right) }{ \left( k+n \right) ^{4} \left( 3\,k+n \right) ^{4}}} \right) $$

Thus this is the log of a rational function of $n$ and $k$. Numerator and denominator of that rational function are both of total degree $2^{m-1}$. I don't see how this could be simplified any further.

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    $\begingroup$ I find it very annoying when the OP completely changes a question after it's answered. $\endgroup$ – Robert Israel Jul 4 '17 at 20:49
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    $\begingroup$ I apologize for doing that, someone pointed out that I should include the full question so that it may give more context, I still greatly appreciate your contribution. $\endgroup$ – JRose Jul 4 '17 at 20:52
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There is a technique for this that has appeared on several occasions on MSE which I am not able to locate at this time. We introduce

$$f(z) = (-1)^m \frac{m!}{n+zk} \prod_{q=0}^m \frac{1}{z-q}.$$

We suppose that $z=-n/k$ is not an integer from the range $[0, m].$ We then obtain

$$\mathrm{Res}_{z=j} f(z) = (-1)^m \frac{m!}{n+jk} \prod_{q=0}^{j-1} \frac{1}{j-q} \prod_{q=j+1}^m \frac{1}{j-q} \\ = (-1)^m \frac{m!}{n+jk} \frac{1}{j!} \frac{(-1)^{m-j}}{(m-j)!} \\ = (-1)^j \frac{1}{nj+k} {m\choose j}.$$

It follows that

$$S = \sum_{j=0}^m \mathrm{Res}_{z=j} f(z)$$

and since residues sum to zero this means that

$$S = -\mathrm{Res}_{z=-n/k} f(z) - \mathrm{Res}_{z=\infty} f(z).$$

Now the residue at infinity is zero since $\lim_{R\to\infty} 2\pi R/R^{m+2} = 0$ or more formally through

$$-\mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z) = - \mathrm{Res}_{z=0} \frac{1}{z^2} (-1)^m \frac{m!}{n+k/z} \prod_{q=0}^m \frac{1}{1/z-q} \\ = - \mathrm{Res}_{z=0} \frac{1}{z^2} (-1)^m \frac{z\times m!}{zn+k} \prod_{q=0}^m \frac{z}{1-qz} \\ = - \mathrm{Res}_{z=0} z^m (-1)^m \frac{m!}{zn+k} \prod_{q=0}^m \frac{1}{1-qz} = 0.$$

This leaves the contribution from $z=-n/k$ and we get

$$-\mathrm{Res}_{z=-n/k} \frac{1}{k} (-1)^{m} \frac{m!}{z+n/k} \prod_{q=0}^m \frac{1}{z-q} \\ = (-1)^{m+1} \frac{m!}{k} \prod_{q=0}^m \frac{1}{-n/k-q} \\ = (-1)^{m+1} \times m! \times k^{m} \prod_{q=0}^m \frac{1}{-n-qk} \\ = m! \times k^{m} \prod_{q=0}^m \frac{1}{n+qk}.$$

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We recall the Melzak's identity $$f\left(x+y\right)=x\dbinom{x+n}{n}\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{f\left(y-k\right)}{x+k},\, x,y\in\mathbb{R},\, x\neq-k $$ where $f $ is an algebraic polynomial up to degree $n $. So taking $f\left(z\right)\equiv1$ and $x=n/k$ we have $$\frac{1}{k}\sum_{j=0}^{m}\dbinom{m}{j}\frac{\left(-1\right)^{j}}{j+n/k}=\color{red}{\frac{1}{n\dbinom{n/k+m}{m}}}.$$

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