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

Question

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.

Answer 1

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 $$ \lim_{n\rightarrow\infty} \frac{\arctan\frac{1}{n}}{\frac{1}{n}} = 1$$

Answer 2

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.