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Does there exist a sequence $a_n$ of complex numbers such that $\sum _{i = 0}^\infty a_n$ converges and the product $\prod _{i = 0}^\infty (1+a_n)$ does not converge to any complex number(not even 0).

And conversely a sequence $a_n$ s.t. $\prod _{i = 0}^\infty (1+a_n)$ converges to a non zero complex number. But $\sum _{i = 0}^\infty a_n$ does not converge.

We all know that if $\sum _{i = 0}^\infty a_n $ converges absolutely iff the corresp product converges absolutely. So i tried alternating serieses like $ a_n = (-1)^n /n $ and $(-1)^n / \sqrt n $ but that did not work. i tried like $ \sum _{i} \sum _ {k} \frac{ e^{2(pi)i/k}}{k} $but i couldn't go furthur. Can anybody help?

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  • $\begingroup$ Plenty of people "can help", if only you explain what are your thoughts. $\endgroup$ – Did May 3 '13 at 9:29
  • $\begingroup$ sorry , my language was very rude $\endgroup$ – rohit May 3 '13 at 11:08
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    $\begingroup$ The question is much more interesting with the remarks in the last paragraph. +1 from me. $\endgroup$ – Did May 3 '13 at 11:41
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$\textrm{(1) $\sum a_n$ converges, $\prod a_n$ does not}$

Take $$\{a_n\}:=\left\{\frac{i}{\sqrt{1}},\frac{-i}{\sqrt{1}},\frac{i}{\sqrt{2}},\frac{-i}{\sqrt{2}},\ldots\right\}$$ Then $$\sum a_n=i\sum(-1)^{n-1}b_n$$ where $$\{b_n\}=\left\{\frac{1}{\sqrt{1}},\frac{1}{\sqrt{1}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\ldots\right\}$$ is a sequence of positive real numbet, non-increasing and infinitesimal, hence by Leibniz the series $$\sum (-1)^{n-1}b_n$$ converges, hence $\sum a_n$ converges too.

Now consider the infinite product $\prod (1+a_n)$. For every positive integer $N$, the $2N$-partial product is $$\prod_{1}^{2N}a_j=\left(1+\frac{i}{\sqrt{1}}\right)\left(1-\frac{i}{\sqrt{1}}\right)\left(1+\frac{i}{\sqrt{2}}\right)\left(1-\frac{i}{\sqrt{2}}\right)\cdot\ldots\cdot\left(1-\frac{i}{\sqrt{2N}}\right)$$

$$=\left(1-\frac{i^2}{1}\right)\left(1-\frac{i^2}{1}\right)\ldots\left(1-\frac{i^2}{2N}\right)$$

$$=\left(1+1\right)\left(1+\frac{1}{2}\right)\ldots \left(1+\frac{1}{2N}\right)$$

$$=2\cdot\frac{3}{2}\cdot\frac{6}{3}\ldots\frac{2N+1}{N}$$

$$=2N+1$$ Thus $$\lim_{N\rightarrow +\infty}\prod_{1}^{2N}(1+a_n)=\lim_{N\rightarrow +\infty} 2N+1=+\infty$$ hence also diverges the following $$\lim_{N\rightarrow +\infty}\prod_{1}^{N} (1+a_n)=\prod_{1}^{+\infty}(1+a_n)$$

$\textrm{(2)$\prod (1+a_n)$ converges, $\sum a_n$ does not}$

Take $\{a_n\}$ defined by $$a_{2n-1}:=\frac{1}{\sqrt{n}}$$ $$a_{2n}:=-\frac{1}{1+\sqrt{n}}$$ for every $n\geq 1$. Then $$\sum_{n=1}^{+\infty}a_n=\sum_{n=1}^{+\infty}a_{2n-1}+\sum_{n=1}^{+\infty}a_{2n}$$

$$=\sum_{n=1}^{+\infty}\frac{1}{\sqrt{n}}-\sum_{n=1}^{+\infty}\frac{1}{1+\sqrt{n}}$$

$$=\sum_{n=1}^{+\infty}\left(\frac{1}{\sqrt{n}}-\frac{1}{1+\sqrt{n}}\right)$$

$$=\sum_{n=1}^{+\infty}\frac{1}{n+\sqrt{n}}$$

But $n+\sqrt{n}<2n$, hence $\frac{1}{n+\sqrt{n}}>\frac{1}{2n}$, and so $$\sum_{1}^{\infty}\frac{1}{n+\sqrt{n}}\geq\frac{1}{2}\sum_{1}^{\infty}\frac{1}{n}$$ and you get divergence by comparison with harmonic series.

Now, as for the divergence of infinite product, consider: $$\prod_{1}^{\infty}(1+a_n)=\prod_{1}^{\infty}(1+a_{2n})\cdot\prod_{1}^{\infty}(1+a_{2n-1})$$

$$=\prod_{1}^{\infty}(1+a_{2n})\cdot (1+a_{2n-1})$$

$$=\prod \left(1-\frac{1}{1+\sqrt{n}}\right)\left(1+\frac{1}{\sqrt{n}}\right)$$

$$=\ldots$$

$$=\prod\frac{\sqrt{n}(1+\sqrt{n})}{\sqrt{n}(1+\sqrt{n})}$$

$$=\prod_{1}^{\infty}1=1$$

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  • $\begingroup$ +1. Some minor but confusing typos in the second half: surely you mean that $a_{2n-1}$ is negative, otherwise the calculations don't make sense? $\endgroup$ – Erick Wong May 3 '13 at 12:15
  • $\begingroup$ @ErickWong thanks, corrected $\endgroup$ – Federica Maggioni May 3 '13 at 12:26
  • $\begingroup$ Why in the series in the example $(2) $ we can change the order ? The terms aren't positive $\endgroup$ – WLOG Feb 4 '14 at 21:31

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