Convergence of series $\sum \arcsin\left(\frac{r_n \sin(\theta_n)}{\sqrt{1+2r_n\cos(\theta_n)+r_n^2}}\right)$

Question

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.

Answer 1

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.

Answer 2

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.