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

How do I evaluate $\displaystyle\lim_{k\to\infty} \prod_{i=1}^{k}(1-\alpha_i+\alpha_i^2)$?

Here, $\alpha_k\in (0,1)$ for every $k\in\mathbb{N}$ and $\displaystyle\lim_{k\to\infty}\alpha_k=0$.

share|cite|improve this question
The result will clearly depend in the sequence... (if you pick any sequence $(\beta_k)_{k\geq1}$ such that all its terms are in $(0,1]$ and such that the product $\prod_{k\geq1}\beta_k$ converges to $P$, there is a unique sequence $(\alpha_k)_{k\geq1}$ with $\alpha_k\in[0,1/2)$, $\beta_k=1-\alpha_l+\alpha_k^2$, $\lim\limits_{k\to\infty}\alpha_k=0$ and therefore $\prod_{k\geq1}(1-\alpha_l+\alpha_k^2)=P$. – Mariano Suárez-Alvarez Feb 6 '12 at 2:26
Moreover, in order for the limit to be nonzero you need $\sum_{n=1}^\infty (\alpha_n - \alpha_n^2)$ to converge. – Robert Israel Feb 6 '12 at 2:42
Thank you for all your interesting comments. Please help me to give a complete solution of this problem. – impartialmale Feb 6 '12 at 2:52
What Robert said is about all you can say without additional information on $a_i$. – anon Feb 6 '12 at 3:00
Dear Robert. Thank you in advance for your comments. I want to know where we can find the fact you said. – impartialmale Mar 17 '12 at 16:57
up vote 3 down vote accepted

The limit of the products $\prod\limits_{i\leqslant k}(1-\alpha_i+\alpha_i^2)$ when $k\to\infty$ is a nonnegative number in $[0,1)$, which is positive if and only if the sum of the series $\sum\limits_k\alpha_k$ is finite.

Edit: The WP page on infinite products might prove useful and, to begin with, the first section on convergence criteria.

Once this is ingested, one can come back to the question here. Consider $\beta_k=\alpha_k-\alpha_k^2$. For every $\alpha_k$ in $(0,\frac12)$, $\beta_k\leqslant\alpha_k\leqslant2\beta_k$. Since $\alpha_k\to0$ by hypothesis, this proves that the series $\sum\limits_k\alpha_k$ converges if and only if the series $\sum\limits_k\beta_k$ does.

share|cite|improve this answer
How can we prove this fact? – impartialmale Mar 17 '12 at 14:59
I am confused: should I answer you here or there? Cruel, terrible, insoutenable dilemne... – Did Mar 17 '12 at 15:40
Thank you in advance for your comments. I am waiting the reply from you. Could you give me some references related to your fact? – impartialmale Mar 17 '12 at 16:52
See Edit. (But to call any of this my fact is inappropriate.) – Did Mar 17 '12 at 17:05

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.