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.
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 $\alpha = 0$.