Convergence of series $\sum \arcsin\left(\frac{r_n \sin(\theta_n)}{\sqrt{1+2r_n\cos(\theta_n)+r_n^2}}\right)$ Let $r_n$ be a sequence of strictly positive numbers converging to $0^+$.  Let $\theta_n$ be a sequence of values in $[0,\tfrac{\pi}{2}]$ which may or may not converge.
Consider the following sequence
$$
\beta_n = \arcsin\left(\frac{r_n \sin(\theta_n)}{\sqrt{1+2r_n\cos(\theta_n)+r_n^2}}\right)
$$
$\beta_n$ is a strictly positive sequence converging to zero. This is the imaginary part of $\log(1 +  r_ne^{i\theta_n})$.
From $\theta\in[0,\tfrac{\pi}{2}]$ it follows that $0\leq \sin(\theta_n),\cos(\theta_n)\leq 1$. Let $\theta = \liminf_{n\to\infty}\sin(\theta_n)$. We then have
$$
0\leq \arcsin\left(\frac{r_n \theta}{1+r_n}\right) \leq \beta_n \leq \arcsin\left(\frac{r_n}{\sqrt{1+r_n^2}}\right).
$$
Since $\arcsin\left(\frac{x}{\sqrt{1+x^2}}\right)\sim x$, and $\arcsin\left(\frac{x\theta}{1+x}\right)\sim x\theta$ we have that
$$
0 \leq \theta \sum r_n \leq \beta_n \leq \sum r_n,
$$
The inequalities being true ``eventually''.
We only care about convergence, so we might as well multiply $\sum r_n$ with a larger/smaller constant.
My question is the following, I want to prove that $\sum \beta_n$ is convergent if and only if $\sum r_n$ is. Now, this certainly not true when $\theta_n = 0$, when $\beta_n \equiv 0$. However, if I add another constraint, that there are infinitely many non-zero $\theta_n$'s, then is the statement true?
If $\liminf_{n\to\infty}\sin(\theta_n) \not=0$, it is true by the above proof. The only problem I have is when $\liminf_{n\to\infty}\sin(\theta_n) = 0$, but I can't find a counterexample, and I believe that the statement is true. We can also assume that all $\theta_n$ are non-zero.
If we allow $\theta_n$ to be both positive and negative around 0, then the statement is false.

This is a supplement to Jacob Manaker's wonderful example.
Assume $\theta_n \ll 1$ and $r_n\ll 1$ then we have that
\begin{align}
\frac{1}{4}r_n\theta_n\le \frac{1}{2}\frac{r_n\theta_n}{1+r_n}\le  \frac{r_n\sin\theta_n}{\sqrt{1+2r_n\cos\theta_n + r_n^2}} \le r_n\theta_n .
\end{align}
In particular, we have that
\begin{align}
\arcsin\left(\frac{1}{2}\frac{r_n\theta_n}{1+r_n} \right) \le \arcsin\left(\frac{r_n\sin\theta_n}{\sqrt{1+2r_n\cos\theta_n + r_n^2}} \right)
\le \arcsin\left(r_n\theta_n \right). 
\end{align}
Notice that
\begin{align}
x\le \arcsin\left(x \right) \le 2x
\end{align}
for $x \in [0, 1]$ then it follows that
\begin{align}
\sum^\infty_{n=1}\arcsin(r_n\theta_n) \sim \sum^\infty_{n=1} r_n\theta_n
\end{align}
and
\begin{align}
\sum^\infty_{n=1}\arcsin(\frac{1}{4}r_n\theta_n) \sim \sum^\infty_{n=1} r_n\theta_n.
\end{align}
Hence, we see that
\begin{align}
\sum^\infty_{n=1}\arcsin\left(\frac{r_n\sin\theta}{\sqrt{1+2r_n\cos\theta_n+r^2_n}} \right)
\end{align}
converges if and only if
\begin{align}
 \sum^\infty_{n=1} r_n\theta_n
\end{align}
converges.
As indicated by Jacob Manaker's example, you could choose $\theta_n$ to cancel out the large contributions coming from $r_n$. So, even if $\sum r_n$ diverges, you could choose $\theta_n$ to make $\sum r_n\theta_n $ converge.
