# Show that $\prod \bigl(1+{i\over n}\bigr)$ diverges.

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.

• Please make sure that the edits I made to your post are okay :) – let's have a breakdown Jun 19 at 10:29
• Yes, it is better now, thanks. – J. Doe Jun 19 at 10:30

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 $$\lim_{n\rightarrow\infty} \frac{\arctan\frac{1}{n}}{\frac{1}{n}} = 1$$
• 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 – J. Doe Jun 19 at 12:20
• @J.Doe Correct. But you can also prove the convergence independently, using $$\log |1+\frac{i}{n}| = \log\sqrt{1+\frac{1}{n^2}}=\frac12 \log(1+\frac{1}{n^2}) \le \frac{1}{2n^2}$$ – Adam Latosiński Jun 19 at 12:38
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.
• Ok, so ${\log(1+{i\over n})\over{i\over n}}\to c$ as $n\to\infty$. How can it help? – J. Doe Jun 19 at 10:26
• So $\sum\log(1+\frac{i}n)$ diverges by the comparison test. – user10354138 Jun 19 at 10:30
• @AdamLatosiński we are comparing the imaginary part, which is the asymptotically dominant part of $\log(1+\frac{i}n)$. – user10354138 Jun 19 at 10:35