# Summation: $\sum \limits_{r=0}^n \frac{ \binom n r}{x+r}$

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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To use \huge is a bad idea. – Did Apr 12 '12 at 11:32
Are you sure about : $\frac 1 {x+r}$? it wasn't $\frac 1 {n+r}$ – Lrrr Apr 12 '12 at 11:32
I am almost sure it's the second one, but my friend is insisting on the first. – Quixotic Apr 12 '12 at 11:45

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.
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 $\Gamma(z+1)$. – Sangchul Lee Apr 12 '12 at 13:20