Prove that the set of accumulation points is circle. Find it center and radius I have the following sequence $a_n= \prod_{k=1}^n (1+\frac{i}{k})$
I need to prove that the set of accumulation points is circle. Also I need to find it center and radius. I don’t know how to attempt, could someone give any hints ? So first I was trying to understand what is the center and radius of circle. Also I think , that I need to somehow rewrite $a_n$’s
 A: We consider the argument and magnitude of the product separately.
The argument of $a_n$ is
$$\sum_{k=1}^n\tan^{-1}\frac1k>\sum_{k=1}^n\frac1k$$
which is the harmonic series, diverging to $\infty$ as $n\to\infty$. Thus the $a_n$ circle around the origin forever.
The magnitude of $a_n$ is
$$\prod_{k=1}^n\sqrt{1+\frac1{k^2}}$$
and
$$\lim_{n\to\infty}a_n=\prod_{k=1}^\infty\sqrt{1+\frac1{k^2}}=\sqrt{\prod_{k=1}^\infty\left(1+\frac1{k^2}\right)}=\sqrt{\frac{\sinh\pi}\pi}$$
(see e.g. here for the derivation of the last step). Thus the accumulation points of $(a_n)$ form a circle centred at the origin with radius $\sqrt{\frac{\sinh\pi}\pi}=1.9173100\dots$
A: You can rewrite $1+\frac{1}{k}$ as $\frac{\sqrt{k^2+1}}{k} e^{i \arctan(1/k)}$. From this, you find
\begin{align*}
a_n = \frac{\prod_{k=1}^n \sqrt{(k^2+1)}}{n!} e^{i \sum_{k=1}^n \arctan(1/k)}
\end{align*}
Now, since $\sum_{k=1}^n \arctan(1/k)$ diverges ($\arctan(x) = \mathcal{O}(x)$ for $x \to 0$), it suffices to show that $\frac{\prod_{k=1}^n \sqrt{(k^2+1)}}{n!}$ converges. Now, $\sqrt{k^2+1} \leq k + \frac{1}{2k}$, and hence, we get $\frac{\prod_{k=1}^n \sqrt{(k^2+1)}}{n!} \leq \prod_{k=1}^n (1 + \frac{1}{2k^2})$, which converges.
