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

Question

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?

Answer 1

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.

Answer 2

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.