Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
"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 ? –  Hardy Mar 12 '12 at 19:00
    
i c the new one thanks –  Hardy Mar 12 '12 at 19:01

1 Answer 1

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 $n$.

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.

share|improve this answer
    
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 ? –  Hardy Mar 13 '12 at 6:47
2  
@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. –  Hardy Mar 13 '12 at 7:05
2  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.