I have been working on this problem for a while and cannot seem to make any progress without coming up with something wrong or hitting a dead end.

Here is what I have so far:

$ \prod (1+a_n) < \infty \implies \sum a_n < \infty $: Similarly we ignore finitely many terms until $|a_n| \leq 1/2$ and we use the taylor series for the product. We have that $\prod 1+ a_n$ converging im plying that $\sum \log (1+a_n)$ converges to a nonzero limit since none of the factors are 0 as $a_n \neq -1$. We have that \begin{eqnarray} |\sum \log(1+a_n) | =| \sum (a_n-a_n^2/2 +\ldots) | \\ \geq | \sum (a_n-|a_n^2/2 +\ldots|) | \geq \left| \sum a_n-|a_n|^2-|a_n|^3-\ldots \right| \\ = \left|\sum a_n-|a_n|^2(1+|a_n|+|a_n|^2+|a_n|^3+\ldots \right| \end{eqnarray} by the triangle inequality. Thus $\infty > |\sum \log(1+a_n) | \geq \left|\sum a_n-2|a_n|^2 \right|$ from the previous part. Thus $\left|\sum a_n-2|a_n|^2 \right|$ is convergent, and since $\sum|a_n|^2$ is absolutely convergent we can split the series (I don't really know if this is even true) and we have that partial sums $|\sum a_n|$ is bounded.

Any help would be appreciated! Also any good references for getting better at this kind of stuff would be great!!

  • $\begingroup$ This is an exercise in Stein and Shakarchi's Complex Analysis, Chapter 5. $\endgroup$ – Jack Apr 5 '13 at 16:17
  • $\begingroup$ It may very well be, my teacher just put it on the problem set. He took problems from lots of different books without citing them, sometimes because he made minor changes. $\endgroup$ – Steven-Owen Apr 5 '13 at 21:23

For every $|a|\lt\frac12$, $0\lt a-\log(1+a)\lt a^2$. Since $\sum\limits_n|a_n|^2$ converges, $|a_n|\lt\frac12$ for every $n$ large enough. Hence $\sum\limits_n\left(a_n-\log(1+a_n)\right)$ converges absolutely as soon as every $\log(1+a_n)$ exists.

This implies that $\sum\limits_na_n$ and $\sum\limits_n\log(1+a_n)$ both converge or both diverge. In particular, if $\sum\limits_n\log(1+a_n)$ converges, then $\sum\limits_na_n$ converges.

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    $\begingroup$ How do you know the part that $\sum |a_n - \log(1+a_n)|$ converges implies that $\sum a_n$ converging and $\sum \log(1+a_n)$ converging are equivalent? Also, thanks! $\endgroup$ – Steven-Owen Apr 15 '12 at 21:32
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    $\begingroup$ @jake: The triangle inequality ensures that $$ \left|\sum a_n-\sum\log(1+a_n)\right|\le\sum\left|a_n-\log(1+a_n)\right| $$ $\endgroup$ – robjohn Apr 15 '12 at 21:39

The statement is not true. Consider e.g. $a_n = -1/(n+1)$, where $\prod_{n=1}^N (1+a_n) = 1/(N+1) \to 0$ as $N \to \infty$.

Note that $\prod_n (1+a_n)$ converges to a nonzero limit iff $\sum_n \log (1 + a_n)$ converges, but you didn't say the limit had to be nonzero.

EDIT: @Didier points out the convention that part of the definition of "convergence" for an infinite product is that the limit is nonzero. This is not universal, however: see e.g. Rudin, "Real and Complex Analysis", sec. 15.1, or Conway, "Functions of One Complex Variable", sec. VII.5.

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    $\begingroup$ About the convergence of infinite products. $\endgroup$ – Did Apr 15 '12 at 21:16
  • $\begingroup$ Sorry, you are correct, I should have added that the product is converging to a non-zero limit. $\endgroup$ – Steven-Owen Apr 15 '12 at 21:25

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