0
$\begingroup$

enter image description here

See the attached about infinite product convergence proof from Stein & Shakarchi, Complex Analysis Lectures (p 141)

As per the proof, if $ \sum | a_n | $ converges, then $ \prod (1 + a_n) $ will also converge to a non zero number as long as none of the terms are zero.

Consider $ a_n = - \frac{1}{2^n} $

Now we have $ \sum | a_n | $ converging to -1. So as per the proof, the corresponding $ \prod (1 + a_n) $ must converge.

On the other hand, all the terms of the product are less than 1. Such infinite product should diverge to zero. (Think $ \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10} \cdots $ )

Which one wins?

$\endgroup$

closed as unclear what you're asking by Did, The Phenotype, José Carlos Santos, A. Goodier, Brian Borchers Mar 29 '18 at 21:29

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ "Such infinite product should diverge to zero. (Think .9∗.9∗...)" – Only if all factors are less then a constant $k < 1$. But here the factors converge to $1$. The product converges to a non-zero value exactly for the reasons you cited above. $\endgroup$ – Martin R Mar 29 '18 at 5:36
  • $\begingroup$ Fun fact: $0.9\times0.9\times\cdots = \prod(1+a_n)$ where $a_n=-0.1$. Does $\sum|a_n|$ converge in this case? $\endgroup$ – Rahul Mar 29 '18 at 5:37
  • $\begingroup$ Hmmm... you guys are right! @Martin-r can you please add as an answer, so I can accept it? Thanks. $\endgroup$ – Shree Mar 29 '18 at 5:50
  • $\begingroup$ I’m confused on how a product diverges to $0$. Would tending towards a concrete real number not he considered convergence? $\endgroup$ – Chase Ryan Taylor Mar 29 '18 at 12:49
  • 1
    $\begingroup$ @ChaseRyanTaylor It's jargon. If $\prod a_j=0$ but $a_j\ne0$ for every $j$ then the product is said to diverge to $0$. (Why would we make such a curious definition? One reason: Say $a_j>0$ for every $j$; we need this definition to be able to say that $\prod a_j$ converges if and only if $\sum\log a_j$ converges.) $\endgroup$ – David C. Ullrich Mar 29 '18 at 16:39
2
$\begingroup$

The infinite $\prod ( 1 - \frac{1}{2^n} )$ is convergent to a non-zero value because the series $\sum \frac{1}{2^n}$ converges and none of the factors is zero.

Your argument

On the other hand, all the terms of the product are less than 1. Such infinite product should diverge to zero. (Think $ \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10} \cdots $ )

would only apply to a product $\prod ( 1 + a_n )$ where all factors are uniformly less than one, i.e. $$ 1 + a_n \le k $$ for some constant $k < 1$. But then $\sum | a_n|$ diverges, so there is no contradiction.

$\endgroup$
2
$\begingroup$

A reasonably trivial example of a telescoping product that converges, yet all its terms are strictly less than unity, is $$a_n = 1 - \frac{1}{n^2}.$$ Then $$\prod_{n=2}^\infty a_n = \lim_{N \to \infty} \prod_{n=2}^N \frac{n-1}{n} \frac{n+1}{n} = \lim_{N \to \infty} \frac{(N-1)!(N+1)!}{2(n!)^2} = \lim_{N \to\infty} \frac{N+1}{2N} = \frac{1}{2}.$$ So clearly there is a problem with your claim that if $0 < a_n < 1$ for all $n$, that $\prod_n a_n \to 0$. In fact, the above calculation demonstrates that the original product $$\prod_{n = 1}^\infty \left( 1 - \frac{1}{2^n}\right) $$ must be greater than $1/4$ and less than $1$, since its terms are bounded below by $1 - 1/n^2$ for all $n \ge 2$, and for $n = 1$, the extra factor of $1/2$ is easily extracted.

$\endgroup$
  • $\begingroup$ I see what you are saying, but I would characterize the root cause as $ a_n $ is not less than 1 as $ n \rightarrow \infty $, rather it converges to 1. $\endgroup$ – Shree Mar 29 '18 at 6:10
  • 1
    $\begingroup$ @Shree I don't understand your comment. Explain why you think $1 - 1/n^2 \not\lt 1$ for $n \ge 2$? As I have defined it, $a_n = 1 - 1/n^2$ is always strictly between $0$ and $1$ for $n \ge 2$, and thus furnishes an example of a sequence of numbers that are all less than $1$, yet their product is not zero--i.e., a counterexample to your flawed reasoning that an infinite product of numbers less than $1$ would somehow "eventually" tend to $0$. $\endgroup$ – heropup Mar 29 '18 at 6:13
  • 1
    $\begingroup$ I am persuaded by Martin’s explanation above viz. The eventual tending to 0 would only apply to a product $ \prod ( 1 + a_n ) $ where all factors are uniformly less than one, i.e. All $ (1+a_n) \le k $ for some constant $ k<1 $. I believe this is not the case for $ 1+a_n = ( 1 - \frac{1}{2^n}) $ or $ 1+ a_n = ( 1 - \frac{1}{n^2}) $ $\endgroup$ – Shree Mar 29 '18 at 6:40

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