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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.

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"Cauchy's Theorem" is like saying "Mozart's composition", "Euler's paper", or "president's speech"... there's so many to choose from – Arturo Magidin Mar 12 '12 at 18:59
That's fair enough, please suggest a more appropriate title ? – Comic Book Guy Mar 12 '12 at 19:00
i c the new one thanks – Comic Book Guy Mar 12 '12 at 19:01
up vote 3 down vote accepted

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 $n$.


$$\sum_{k=1}^{n} a_k^2b_k - \sum_{k=1}^{n} \log(1+a_k) = \sum_{k=1}^{n} a_k$$

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.


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.

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Firstly buddy thanks very much for your answer. I have to tell you i think it is quite ingenious how you created a $b_{n}$ to represent the problem. I would love to hear why you chose that particular value for $b_{n}$. Although i am not sure that we are allowed to choose $a_{n}$ goes to zero as the result states a conditional convergence only instead of an absolute one. Any thoughts ? – Comic Book Guy Mar 13 '12 at 6:47
@Hardy: When considering products, it is natural to consider the logarithms and trying to get $a_n$, $a_n^2$ and $\log(1+a_n)$ into the picture kind of leads to $b_n$. It could also be that I might have read a similar proof earlier... Conditional convergence implies convergence (not necessarily absolute) and so $a_n \to 0$ isn't it? – Aryabhata Mar 13 '12 at 6:52
see i think absolute convergence could only occur if the underlying sequence $a_{n}$ goes to zero, on the contrary conditional convergence can occur i believe even with the underlying sequence $a_{n}$'s even and odd terms going to $+\inf$ and $-\inf$. I can n't think of a definitive example of the top of my head but i am quite confident that it exists. If you are not convinced please let me know and i shall find an example of such a series for our discussion. – Comic Book Guy Mar 13 '12 at 7:05
@Hardy: If $S_n =\sum_{k=1}^n a_k$ is convergent (absolute or not), then $a_n = S_n - S_{n-1} \to 0$. Note that the sum of positive terms could go to $\infty$. We are talking about the individual term. For instance $\sum \frac{(-1)^n }{n}$ is the classic example of such a series and what $a_n \to 0$ is saying is that $\frac{(-1)^n }{n} \to 0$. – Aryabhata Mar 13 '12 at 7:32

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