Proof verification: $\prod_{k=1}^{\infty} (1+\frac{i}{k^2}) \notin \mathbb R$. Show that  $$\mathcal Z=\prod_{k=1}^{\infty} (1+\frac{i}{k^2}) \notin \mathbb R,$$ where $i=\sqrt{-1}$.
Here is my attempt, please check it and comment if it is correct or if there are any logical gaps or if there's an easier method for any/every part of it.
Proof:
Let $\arg(z)$ denote the principal argument of the complex number $z$.
$z \in \mathbb R$ iff $\arg(z)=0$ or $\arg(z)=\pi$.
Lemma 1. If $z_1=a+ib, z_2=c+id$ for $a,b,c,d\in\mathbb R^+,$ $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$
Proof:
$ \arg(z_1)+\arg(z_2)=\arctan\left(\frac b a\right) +\arctan\left( \frac d c\right)=\arctan\left(\frac{bc+ad}{ac-bd}\right)$
$z_1z_2=(ac-bd)+i(ad+bc) \implies \arg(z_1z_2)=\arctan\left(\frac{bc+ad}{ac-bd}\right)$.
Lemma 1 $\implies \arg\left(\prod_{r=1}^n z_r\right)=\sum_{r=1}^n \arg(z_r)$.
$\implies \arg(\mathcal Z)=\sum_{k=1}^{\infty} \arg(1+\frac{i}{k^2})=\sum_{k=1}^{\infty} \arctan\left(\frac{1}{k^2}\right)$.
Now $\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$
$\implies \arctan\left(\frac{1}{k^2}\right)=\frac{1}{k^2}-\frac{1}{3k^6}+\frac{1}{5k^{10}}-\frac{1}{7k^{14}}+\frac{1}{9k^{18}}-\cdots$
But then,
$\frac{1}{3k^6} \gt \frac{1}{5k^{10}}$
$\frac{1}{7k^{14}} \gt \frac{1}{9k^{18}}$ and so on...
$\implies \arctan\left(\frac{1}{k^2}\right)=\frac{1}{k^2}- t$ for some $t\gt 0$
$\implies \arctan\left(\frac{1}{k^2}\right)\lt \frac{1}{k^2}$
$\implies \sum_{k=1}^{\infty} \arctan\left(\frac{1}{k^2}\right) \lt \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$
and since $\frac{1}{k^2} \gt 0, \arctan\left(\frac{1}{k^2}\right)\gt 0\implies \sum_{k=1}^{\infty} \arctan\left(\frac{1}{k^2}\right) \gt 0$
$$\implies 0\lt \sum_{k=1}^{\infty} \arctan\left(\frac{1}{k^2}\right) \lt \frac{\pi^2}{6} \lt \pi$$
$$\implies \arg(\mathcal Z) \in (0,\pi) \implies \mathcal Z \notin \mathbb R.$$
 A: I'll try to prove this evaluating the closed form in terms of gamma function. Starting with partial products note that
$$\prod\limits_{k=1}^{n} (x+k)= \frac{\Gamma (1+x+n)}{\Gamma (x+1)}$$
Hence we have
$$P=\prod\limits_{k=1}^{\infty } \left(1+\frac{i}{k^2}\right)=\lim_{n\to \infty } \prod\limits_{k=1}^{n} \frac{(k+i\sqrt{i})(k-i\sqrt{i})}{k^2}$$
Using the earlier mentioned identity we get :
$$P=\lim_{n\to \infty } \frac{\Gamma( n+i\sqrt{i}+1)\Gamma(n-i\sqrt{i}+1)}{\Gamma(1+i\sqrt{i})\Gamma(1-i\sqrt{i})\Gamma^2(n+1)}$$
Now we use the fact that :
$$\lim_{n\to \infty } \frac{\Gamma(n+a)}{\Gamma(n+b)}=n^{a-b}$$
Hence we get
$$P=\lim_{n\to \infty } n^{1+i\sqrt{i}+1-i\sqrt{i}-1-1}\cdot \frac{1}{\Gamma (1+i\sqrt{i})\Gamma(1-i\sqrt{i})}$$
The powers cancel out hence we get :
$$P=\frac{1}{\Gamma (1+i\sqrt{i})\Gamma(1-i\sqrt{i})}$$
It doesn't belong to $\mathbb{R}$ follows from the fact that $$\Gamma (1+z) = \Gamma (z)\cdot z$$ and using this in the formula $$\Gamma (z)\Gamma (1-z)=\frac{\pi }{\sin ( \pi z)}$$. Clearly $z=i\sqrt{i}$ Plugging in the value you'll get the result in terms of $\csc (i\sqrt{i}\pi)$ which indeed can/does not belong to $\mathbb{R}$
I hope I was clear. 
Thank you !)
