Convergence of a Taylor series expansion involving $\tan^{-1}$

Question

Taylor series expansion of $\tan^{-1}\left(\cfrac{\sin(\theta)+r\cos(\phi)}{\cos(\theta)+r\sin(\phi)}\right)$ around $r=0$ is as follows:

$$\begin{align}\tan^{-1}\left(\frac{\sin(\theta)+r\cos(\phi)}{\cos(\theta)+r\sin(\phi)}\right)={} &\theta+r\cos(\psi)-\frac{r^2}{2} \sin(2\psi)-\frac{r^3}{3}\cos(3\psi)+\frac{r^4}{4}\sin(4\psi)\\ &{}+\frac{r^5}{5}\cos(5\psi)-\frac{r^6}{6}\sin(6\psi)-\frac{r^7}{7}\cos(7\psi)+\frac{r^8}{8}\sin(8\psi)+\cdots\\ \end{align}\tag{1}$$ where $\psi\equiv\theta+\phi$.

I am interested in the radius of convergence of this series for $\psi\in[0,2\pi]$. Clearly, it converges for $|r|<1$: since $\sin x\in[-1,1]$ and $\cos x\in [-1,1]$ we can replace the trigonometric functions by their upper and lower bounds to yield the following upper and lower bounds for the RHS of (1): $\theta\pm\sum_{i=1}^\infty\frac{r^i}{i}=\theta\pm\log(1-r)$. However, I am wondering what happens to the series when $|r|\geq1$. Any hints/tips/ideas on how to evaluate this case?

Answer 1

HINT:

For $|r|>1$, we have

$$\begin{align} \limsup_{n\to \infty}\sqrt[n]{\left|\frac{(-1)^nr^{2n}\sin(2n\psi)}{2n}\right|}&=r^2\,\limsup_{n\to \infty}\sqrt[n]{\left|\frac{\sin(2n\psi)}{2n}\right|}\\\\ &=r^2\\\\ &>1 \end{align}$$

with an analogous result for the odd terms of the series.

For $r=1$, the series convergence is guaranteed using Leibniz's Test.

For $r=-1$, the series convergence is guaranteed for $\psi \ne 0$ by Dirichlet's Test

Answer 2

Since Dr. MV already answered the question for $|r|>1$, this is probably too long for a comment.

I may be wrong but it seems to me that the Taylor series expansion around $r=0$ is $$\tan^{-1}\left(\frac{\sin(\theta)+r\sin(\phi)}{\cos(\theta)+r\cos(\phi)}\right)=\theta+\sum_{n=1}^\infty \frac{(-1)^n}{n}\sin\left(n(\theta-\phi)\right)\,r^n=\theta+\sum_{n=1}^\infty \frac{(-1)^n}{n}\sin\left(n\psi\right)\,r^n$$ with $\color{red}{\psi=\theta-\phi}$.

Consider $\theta=\phi$ in the expression and in the one you propose.