Well it is relatively well known that the condition for absolute convergence is given by the following theorem: In order that the infinite product $\prod _{n=1}^{\infty }\left( 1+a_{n}\right) $ may be absolutely convergent, it is necessary and sufficient that the series $\sum _{n=1}^{\infty }a_{n}$ should be absolutely convergent.
I am trying to prove a little less famous result from Cauchy, which states If $\sum _{n=1}^{\infty }a_{n}$ be a conditionally convergent series of real terms, then $\prod _{n=1}^{\infty }\left( 1+a_{n}\right) $ converges (but not absolutely) or diverges to zero according as $\sum _{n=1}^{\infty }a_{n}^{2}$ converges or diverges.
Some thoughts towards the Proof Although i could eb wrong here but since we do not know that $a_{n}\rightarrow 0$ under the given circumstances i guess a proof by comparison to $\sum _{k=0}^{\infty }\dfrac {1} {k^{2}}$ or which required the ln series kind of seem to fall apart. I was hoping some one could possibly provide an idea/ strategy for this proof.
Any help would be much appreciated.
We can assume $\displaystyle a_n \neq 0$.
Define $\displaystyle b_n$ as follows
$$ b_n = \frac{\log(1+a_n) - a_n}{a_n^2}$$
Notice that $\displaystyle b_n \lt 0$ for all $n$.
Thus
$$\sum_{k=1}^{n} a_k^2b_k - \sum_{k=1}^{n} \log(1+a_k) = \sum_{k=1}^{n} a_k$$
Using the Taylor expansion of $\displaystyle \log (1+x)$, (note that $\displaystyle a_n \to 0$), we see that $\displaystyle b_n \to \frac{-1}{2}$.
Now, it is well known that if $\displaystyle \sum x_n$ converges absolutely and $\displaystyle y_n$ is bounded, then $\displaystyle \sum x_n y_n$ converges.
Thus
If $\displaystyle \sum_{k=1}^{n} a_k^2 $ converges then so does $\displaystyle \sum_{k=1}^{n} a_k^2 b_k$ and as a consequence, so does $\displaystyle \sum \log (1+a_k)$ and $\displaystyle \prod (1+a_k)$.
If $\displaystyle \prod(1+a_n)$ converges, then so does $\displaystyle \sum_{k=1}^{n} \log(1+a_k)$, and so $\displaystyle \sum_{k=1}^{n} a_k^2 b_k$ converges. Since $\displaystyle b_k \lt 0$, this convergence is absolute and thus the sequence $\displaystyle \sum_{k=1}^{n} a_n^2 b_k \times \frac{1}{b_k} = \sum_{k=1}^{n} a_k^2$ converges, as the sequence $\displaystyle \frac{1}{b_n}$ is bounded.