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I have a sequence of complex numbers $a_1,a_2,...$ such that $a_i \neq -1$. I then have the infinite product $\prod\limits_{n=1}^{\infty}(1+a_n)$ which I know converges to a non zero complex number. I was wondering if this is enough to guarantee the convergence of $\sum\limits_{n=1}^\infty\frac{a_n}{1+a_n}$.

I couldn't come up with an obvious counterexample and my first idea for proving it was assuming $\prod\limits_{n=1}^{\infty}(1+a_n)=S$, so $\sum\limits_{n=1}^{\infty}\ln(1+a_n)=\ln S$ converges but have no idea how to deal with this since $1+a_i$ is complex.

If this diverges in general, I was wondering if there are any necessary and sufficient conditions for the sum to converge such as $\sum\limits_n|a_n|<\infty$ or that the $a_n$ are positive real numbers.

Any help would be awesome,and thanks for taking a look!

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  • $\begingroup$ I believe $\sum \ln(1+a_n) $ converging implies $\sum a_n$ converges, even for complex numbers, but I'm still looking for the proof (@twakhare it was, people keep giving diverging to 0 products). $\endgroup$
    – user24142
    Aug 11, 2015 at 0:49

2 Answers 2

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The infinite product $\prod_n (1+a_n)$ converges (to a nonzero value) if and only if $\sum_n \log(1+a_n)$ converges (for a branch of $\log$ that is analytic in a neighbourhood of $1$, with $\log(1) = 0$). This does not in general imply $\sum_n \dfrac{a_n}{1+a_n}$ converges. For example, take a positive sequence $b_k \to 0 + $ and let $c_k = 1/(1+b_k) - 1$. Thus $\log(1+c_k) = -\log(1+b_k)$. On the other hand, $$ \dfrac{c_k}{1+c_k} + \dfrac{b_k}{1+b_k} = \dfrac{-b_k^2}{1+ b_k} \ne 0$$ Consider a sequence $a_n$ that, for each $k$, repeats $b_k, c_k$ enough times for the partial sum of $a_n/(1+a_n)$ to change by $1$, before progressing to the next $k$. Then $\sum \log(1+a_n)$ converges to $0$ but $\sum a_n/(1+a_n)$ diverges.

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    $\begingroup$ Very nice. A simple sequence constructed using this idea is $a_n =\{\color{red}{\frac{1}{2},-\frac{1}{3},\frac{1}{2},-\frac{1}{3}}$, $\color{blue}{\frac{1}{3},-\frac14,\frac13,-\frac14,\frac13,-\frac14}$, $\color{green}{\frac14,-\frac15},\color{green}{\frac14,-\frac15},\color{green}{\frac14,-\frac15},\color{green}{\frac14,-\frac15},\ldots\}$. Then $\prod(1+a_n) = \color{red}{1}\cdot \color{blue}{1}\cdot \color{green}{1}\cdots$ while $\sum\frac{a_n}{1+a_n} = -\left(\color{red}{\frac{1}{3}} + \color{blue}{\frac{1}{4}} + \color{green}{\frac{1}{5}} + \ldots\right)$. $\endgroup$
    – Winther
    Aug 11, 2015 at 1:36
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It's OK for absolute convergence. But not in general.

Take $\log(1+a_n) = (-1)^n/\sqrt{n}$ for $n \ge 1$, so $\sum \log(1+a_n) = \sum (-1)^n/\sqrt{n}$ converges (conditionally). But $$ \frac{a_n}{1+a_n} = \frac{(-1)^n}{\sqrt{n}} - \frac{1}{2n} +O\left(\frac{1}{n^{3/2}}\right) $$ and thus $\sum a_n/(1+a_n)$ diverges.

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