Common denominator example

Question

Assuming sum: $$\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k+\alpha+1} =\frac{1}{0!(\alpha+1)}-\frac{n}{1!(\alpha+2)} + \frac{n(n-1)}{2!(\alpha+3)}+\cdots+\frac{n!(-1)^n}{n!(\alpha+n+1)}$$ How does one derive generally, that this can be rewritten as: $$\frac{n!}{(\alpha+1)(\alpha+2)(\alpha+3)\cdots(\alpha+n+1)}$$ I've got to derive this for smaller $n$, maybe $n=2,3,\ldots$ But i can't seem to get to derive it generally for all $n$. I don't think mathematical induction is the way here. Would someone help me out please?

Answer 1

$$\begin{aligned} \sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k+\alpha+1} & = \sum_{k=0}^n \binom{n}{k} \int_0^1 (-1)^kx^{k+\alpha}\,{dx} \\&= \int_0^1 \sum_{k=0}^n \binom{n}{k}(-1)^kx^{k+\alpha} \,{dx} \\&= \int_{0}^{1}x^{\alpha}(1-x)^{n}\,{dx}\end{aligned} $$

Updated: Denoting the integral by $I_{(\alpha, n)}$ and integrating by parts once, we get

\begin{aligned}I_{(\alpha, n)} & = \int_{0}^{1}{x^{\alpha}}{(1-x)^{n}}\;{dx} =\frac{1}{\alpha+1} \int_{0}^{1}\left(x^{\alpha+1}\right)'(1-x)^{n}\;{dx} \\& = \frac{1}{\alpha+1}\cdot x^{\alpha+1}{(1-x)^{n}}\bigg|_{0}^{1}+\frac{n}{\alpha+1}\int_{0}^{1}x^{\alpha+1}{(1-x)^{n-1}}\;{dx} \\& = \frac{n}{\alpha+1}\int_{0}^{1}x^{\alpha+1}{(1-x)^{n-1}}\;{dx} = \frac{n}{\alpha+1}~I_{(\alpha+1, n-1)}. \end{aligned}

So now we have a recursive formula.

$$\displaystyle \begin{aligned} I_{(\alpha, n)} & = \frac{n}{\alpha+1}~I_{(\alpha+1, n-1)} \\& = \frac{n}{\alpha+1} \cdot \frac{n-1}{\alpha+2}~I_{(\alpha+2, n-2)} \\&= \cdots \\&= \frac{n(n-1)(n-2)\cdots 1}{(\alpha+1)(\alpha+2)\cdots(\alpha+n-1)}I_{(\alpha+n-1, 1)} \end{aligned} $$

But $ \displaystyle I_{(\alpha+n-1,1)} = \int_{0}^{1}{x^{\alpha+n-1}}{(1-x)}\;{dx} = \frac{1}{(\alpha+n)(\alpha+n+1)}$ thus $$ \begin{aligned} I_{(\alpha, n)} & = \frac{n(n-1)(n-2)\cdots 1}{(\alpha+1)(\alpha+2)\cdots(\alpha+n+1)} \\& = \frac{n!}{(\alpha+1)(\alpha+2)(\alpha+3)\cdots(\alpha+n+1)} \end{aligned} $$

Answer 2

$$ \frac{n!}{(\alpha+1)\ldots(\alpha+n+1)}$$ is a rational function $F(\alpha)$ with simple poles at $\alpha = -1, \ldots, \alpha = -n-1$ and nowhere else, and $F(\infty) = 0$. Therefore it has a partial fraction expansion of the form $$F(\alpha) = \sum_{k=1}^{n+1} \frac{c_k}{\alpha + k }$$ Now you just need to find the residues at each pole: $c_k = \text{Res}(F(\alpha); \alpha = -k)$. You may use the formula $$ \text{Res}\left(\frac{g(z)}{z-p}; z=p\right) = g(p)$$