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

J. M. ain't a mathematician
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impartialmale
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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-Álvarez Feb 06 '12 at 02:26
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    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 06 '12 at 02:42
  • Thank you for all your interesting comments. Please help me to give a complete solution of this problem. – impartialmale Feb 06 '12 at 02:52
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    What Robert said is about all you can say without additional information on $a_i$. – anon Feb 06 '12 at 03:00
  • 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
  • Deat Robert. Can you give me a proof of your statement? – impartialmale Mar 17 '12 at 21:40

1 Answers1

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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.

Did
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