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

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?

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.

• Also see here for a counterexample showing if $a_n$ changes sign infinitely often then $\sum a_n$ and $\prod(1+a_n)$ need not converge and diverge together.
– RRL
Commented Nov 5, 2018 at 16:56