Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question

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 $\alpha = 0$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.