Convergence of infinite product $\prod(1+a_n)$ where $a_n$ changes sign

Question

I know that an infinite product $\prod_{n=1}^\infty (1+a_n)$ with $a_n \geq0$ for all $n$ converges if and only if the series $\sum_{n=1}^\infty a_n$ converges. I can prove this using the inequality $e^x \geq 1+x$ which gives

$$\sum_{n=1}^N a_n \leq \prod_{n=1}^N(1+a_n) \leq \exp\left({\sum_{n=1}^N a_n}\right)$$

The left inequality here relies on the $a_n$ being of the same sign.

There is also a theorem that the product converges if and only if the series converges when $-1 < a_n < 0$ for all $n$. The proof also uses the fact that the $a_n$ are all of the same sign.

My question is are there general convergence theorems for infinite products where $a_n$ is of alternating sign or changing sign infinitely often less frequently? If so, how would it be proved?

Answer 1

The theorems you describe only consider products of the form $\prod(1+a_n)$ with $a_n \geqslant 0$ and $\prod(1-b_n)$ with $0\leqslant b_n < 1$.

In general, we have infinite products where $a_n$ could be any real or complex number. The strongest condition that guarantees convergence of $\prod(1+a_n)$ is, of course, absolute convergence. This reverts back to your first theorem.

The product $\prod(1+a_n)$ where $a_n \in \mathbb{R} \text{ or } \mathbb{C}$ is said to be absolutely convergent if $\prod (1+|a_n|)$ is convergent. The product converges absolutely if and only if $\sum|a_n|$ is convergent. Also, convergence of $\prod (1+|a_n|) $ implies convergence of $\prod (1 +a_n)$.

In the absence of absolute convergence, the product may still converge conditionally but, as far as I know, there are no all-encompassing necessary and sufficient conditions.

There are some ad hoc theorems for conditional convergence of products where the sign of $a_n$ is unrestricted. For example:

If the series $\sum a_n^2$ is convergent, then the product $\prod(1+a_n)$ and the series $\sum a_n$ either both converge or both diverge.

To prove this note that if $\sum a_n^2$ converges then $|a_n| \to 0$ and for sufficiently large $n$ we have $|a_n| < 1/2$ and

$$|\log(1+a_n) - a_n| = \left|\sum_{k=2}^\infty(-1)^{k}\frac{a_n^k}{k} \right| \leqslant \frac{a_n^2}{2}\sum_{k=0}^\infty|a_n|^k = \frac{a_n^2}{2}\frac{1}{1-|a_n|}< a_n^2$$

Thus the series $\sum [\log(1+a_n) - a_n]$ is absolutely convergent by the comparison test, and we have existence of the limit,

$$\lim_{N\to \infty} \log \left(\prod_{n=1}^N (1+a_n)\right) - \sum_{n=1}^Na_n =\lim_{N\to \infty} \prod_{n=1}^N [\log(1+a_n)- a_n ] $$

proving that the product and series must converge or diverge together.