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.

No; you need the liminf.
Let $$r_n=\begin{cases}
\frac{1}{n} & 2\mid n \\
\frac{1}{n^2} & 2\nmid n
\end{cases}$$ and $$\theta_n=\begin{cases}
0 & 2\mid n \\
\frac{1}{n} & 2\nmid n
\end{cases}$$  Then $$\beta_n=\begin{cases}
0 & 2\mid n \\
\sin^{-1}{\left(\frac{\sin{\frac{1}{n}}}{\sqrt{n^2+1+2\cos{\frac{1}{n}}}}\right)} & 2\nmid n
\end{cases}$$ $\sum_n{r_n}$ does not converge, but since $\beta_n\sim\frac{1}{n^2}$ for large odd $n$, the sum $\sum_n{\beta_n}$ converges.
