# Does infinite product $\prod ( 1 - \frac{1}{2^n} )$ diverge to 0 or converge [closed]

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?

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

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• "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. – Martin R Mar 29 '18 at 5:36
• 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? – Rahul Mar 29 '18 at 5:37
• Hmmm... you guys are right! @Martin-r can you please add as an answer, so I can accept it? Thanks. – Shree Mar 29 '18 at 5:50
• I’m confused on how a product diverges to $0$. Would tending towards a concrete real number not he considered convergence? – Chase Ryan Taylor Mar 29 '18 at 12:49
• @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.) – David C. Ullrich Mar 29 '18 at 16:39

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.

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.
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.
• 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. – Shree Mar 29 '18 at 6:10
• @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 $0$. – heropup Mar 29 '18 at 6:13
• 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 $1+ a_n = ( 1 - \frac{1}{n^2})$ – Shree Mar 29 '18 at 6:40