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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

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


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

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




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?


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1 Answer 1

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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Correct. It is in my question. The point is for $\alpha<1$. –  Jon Jan 21 '12 at 19:02
@Jon The hypergeometric function will appear in the Borel regularization, which give the formula above, alas. Thus it seems that for $0<\alpha < 1$ and for $\alpha<0$ the sum can be attributed a finite value. –  Sasha Jan 21 '12 at 20:18
Fine. This should be so as I was able to get the result through another route. Of course, if I would be able to match these the result would become generally useful. By the way, I am writing a paper and I aim to put your name on the acknowledgements. Hope this will not hurt you. –  Jon Jan 21 '12 at 20:25
@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 $\alpha \not=0$. –  Sasha Jan 21 '12 at 20:49

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