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How to evaluate $ \sum \limits_{r=0}^n \large \frac{\binom n r} {x+r} $

I got this problem from a friend according to him, $ \binom n r$ is the coefficient of $(1+x)^n$. I am not sure how to approach this one or if it has a nice closed form. Any ideas?

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    I am almost sure it's the second one, but my $f$riend is insisting on the first.2012-04-12

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Let $f(u)=\sum_{r=0}^n{{n\choose r}\over x+r}u^{x+r}$ Then f'(u)=\sum_{r=0}^n{n\choose r}u^{x+r-1}=u^{x-1}\sum_{r=0}^n{n\choose r}u^r=u^{x-1}(1+u)^n So the sum can be expressed as $\int_0^1u^{x-1}(1+u)^n\,du$ Maybe it has a nice expression in terms of beta functions.

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    So far I was just able to obtain an expression in terms of incomplete beta function, though its close relative has a more tractable closed form $\sum_{r=0}^{\infty} \binom{\alpha}{r} \frac{(-1)^{r}}{x+r} = \frac{\pi}{\sin \pi x}\binom{\alpha}{-x},$ where $\alpha, x \in \mathbb{R}$ with $\alpha \geq 0$ and $z!$ is identified with the gamma function $\Gamma(z+1)$.2012-04-12