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Show that $$A:=\prod_{n=1}^\infty \left(1+{i\over n}\right)$$ diverges, where $i$ is the imaginary unit

My attampt:

I've shown that $\displaystyle B:=\prod_{n=1}^\infty\left|1+{i\over n}\right|$ converges. Also, one can notice that $A$ diverges iff $\displaystyle \sum_n\log\left(1+{i\over n}\right)$ diverges.

Notice that $$ \log\left(1+{i\over n}\right) = \underbrace{\left({1\over 2n^2}-{1\over 4n^4}+{1\over 6n^6}-\cdots\right)}_U+i\underbrace{\left({1\over n}-{1\over 3n^3}+{1\over 5n^5}-\cdots\right)}_V $$ We know that both $U$ and $V$ converge where $n>1$. I'm not sure how to continue. Thanks.

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  • $\begingroup$ Please make sure that the edits I made to your post are okay :) $\endgroup$ – let's have a breakdown Jun 19 at 10:29
  • $\begingroup$ Yes, it is better now, thanks. $\endgroup$ – J. Doe Jun 19 at 10:30
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You have $$ \log (1+\frac{i}{n}) = \log |1+\frac{i}{n}| + i\arg (1+\frac{i}{n}) = \log |1+\frac{i}{n}| + i\arctan \frac{1}{n} $$ Series $ \sum_{n=1}^\infty \log |1+\frac{i}{n}|$ is convergent (which is related to the convergence of $B$), so we need to show that $$ \sum_{n=1}^\infty \arctan \frac{1}{n}$$ is divergent, and that can be done with comaprison test to $\sum_{n=1}^\infty \frac{1}{n}$, because $$ \lim_{n\rightarrow\infty} \frac{\arctan\frac{1}{n}}{\frac{1}{n}} = 1$$

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  • $\begingroup$ I want to ensure: You wrote that $\prod \log |1+{i\over n}|$ is convergent because $B= \prod |1+{i\over n}|$ is convergent. The explanation to this is that the $\log$ function is continuous in $\mathbb{R}^+$, am I right? @Adam $\endgroup$ – J. Doe Jun 19 at 12:20
  • $\begingroup$ @J.Doe Correct. But you can also prove the convergence independently, using $$\log |1+\frac{i}{n}| = \log\sqrt{1+\frac{1}{n^2}}=\frac12 \log(1+\frac{1}{n^2}) \le \frac{1}{2n^2} $$ $\endgroup$ – Adam Latosiński Jun 19 at 12:38
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You know $\displaystyle\log\left(1+\frac{i}{n}\right) \sim\frac{i}n$ (as $n\to\infty$) and $\displaystyle\sum\frac{1}{n}$ diverges.

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  • $\begingroup$ Ok, so ${\log(1+{i\over n})\over{i\over n}}\to c$ as $n\to\infty$. How can it help? $\endgroup$ – J. Doe Jun 19 at 10:26
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    $\begingroup$ So $\sum\log(1+\frac{i}n)$ diverges by the comparison test. $\endgroup$ – user10354138 Jun 19 at 10:30
  • $\begingroup$ @user10354138 There's no comparison test for series with complex values. You can only use it for series with real, positive elements. $\endgroup$ – Adam Latosiński Jun 19 at 10:33
  • $\begingroup$ @AdamLatosiński we are comparing the imaginary part, which is the asymptotically dominant part of $\log(1+\frac{i}n)$. $\endgroup$ – user10354138 Jun 19 at 10:35

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