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I have been looking in the known literature before to ask this question that could have a very easy answer. Let me state the problem. I have a series like this

$(1-x)^\alpha= \sum_{n=0}^\infty\left(\begin{array}{c} \alpha \\ n \end{array}\right)(-1)^{n}x^n$

that exists provided $|x|<1$. But I am interested to evaluate this series when it diverges. Is this summable? So, I can get by derivation the following series

$S_1= \sum_{n=0}^\infty\left(\begin{array}{c} \alpha \\ n \end{array}\right)(-1)^{n}n$

and

$S_2= \sum_{n=0}^\infty\left(\begin{array}{c} \alpha \\ n \end{array}\right)(-1)^{n}n(n-1)$

that can be obtained by deriving the preceding one and evaluating them to $x=1$ where the original function just goes to infinity.

$S_1$ and $S_2$ appear to be not summable for all $\alpha>0$. Indeed, if I use Abel summation method I get

$S_1(\epsilon)=-\alpha\frac{(1-e^{-\epsilon})^a}{-1+e^\epsilon}$

and

$S_2(\epsilon)=\alpha(\alpha-1)\frac{(1-e^{-\epsilon})^a}{(-1+e^\epsilon)^2}.$

From Abel summation we can see that $S_1=0$ for $\alpha>1$ and $S_2=0$ for $\alpha>2$ and are infinite otherwise (excluding integers 1 and 2) for $\epsilon\rightarrow 0$.

My question is simple: Is Abel summation the last word for $a<1$? On Hardy's book there are cited some techniques with hypergeometric functions. Are there applicable here and how?

Thanks.

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Clearly: $ \sum_{n=0}^\infty \binom{\alpha}{n} (-1)^n n x^n = x \frac{\mathrm{d}}{\mathrm{d} x} (1-x)^\alpha = -\alpha x (1-x)^{\alpha -1} $ The limit $x \uparrow 1$ exists when $\alpha \geqslant 1$, or, trivially when $\alpha = 0$.

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    @Jon Thanks you, it can not hurt :) My second comment above is full of typos, sorry. I was saying that it seems that the sum can be regularized to a finite value, it seems, for \alpha<1 and $\alpha \not=0$.2012-01-21