Evaluate:$\sum_{n=2}^{\infty}\frac{\tan \theta_{n}}{3^n\left(3-\tan^2\theta_{n}\right)}$

Question

Evaluate:$$\sum_{n=2}^{\infty}\frac{\tan \theta_{n}}{3^n\left(3-\tan^2\theta_{n}\right)}$$

where $$\theta_{n}=\frac{\theta}{3^n}$$ and $0<\theta<\pi$

I did try to find relation between $\tan 3x$ and $\tan x$

$$\tan 3x-3\tan x=\frac{8\tan^3x}{1-3\tan^2x}$$

Also $$3\tan 3x-\tan x=\frac{8\tan x}{1-3\tan^2x}$$

so as to create some kind of telescopic action but couldn't split the expression given in the question

Answer 1

Note that

$$ \frac{\tan x}{3-\tan^2 x} = \frac{\cot x - 3\cot(3x)}{8}. $$

So we have

$$ \frac{\tan \theta_n}{3^n(3-\tan^2 \theta_n)} = \frac{1}{8}\left( \frac{\cot \theta_n}{3^n} - \frac{\cot \theta_{n-1}}{3^{n-1}} \right). $$

Summing this from $n = 2$ to $\infty$ yields a telescoping series with the value

$$ \sum_{n=2}^{\infty} \frac{\tan \theta_n}{3^n(3-\tan^2 \theta_n)} = \frac{1}{8}\left( \lim_{N\to\infty} \frac{\cot \theta_N}{3^N} - \frac{\cot \theta_1}{3} \right) = \frac{1}{8}\left( \frac{1}{\theta} - \frac{\cot (\theta/3)}{3} \right). $$