# Calculating the limit $\lim\limits_{k\to\infty} \prod\limits_{i=1}^{k}(1-\alpha_i+\alpha_i^2)$.

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 $\displaystyle\lim_{k\to\infty}\alpha_k=0$.

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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
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

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.
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.