# Infinite Series (Telescoping?)

$$\sum_{n=0}^\infty \frac{\tan(a/2^n)}{2^n},$$

where $a$ isn’t a multiple of $\pi$. I’ve been going through several telescoping questions, and It seems I have hit a brick wall with this one, any help will be appreciated.

#### Solutions Collecting From Web of "Infinite Series (Telescoping?)"

Since
$$\tan\left(\frac{a}{2^n}\right)\sim_\infty\frac{a}{2^n}$$
then
$$\frac{\tan(a/2^n)}{2^n}\sim_\infty\frac{a}{4^n}$$
so the given series is convergent.

Added (If you ask for the sum) We have
$$\tan t= \frac{1}{\tan t}-\frac{2}{\tan (2t)}$$
hence we find
$$\frac{\tan(a/2^n)}{2^n}= \frac{1}{2^n\tan (a/2^n)}-\frac{1}{2^{n-1}\tan (a/2^{n-1})}=u_n-u_{n-1}$$
where
$$u_n=\frac{1}{2^n\tan (a/2^n)}$$
so by telescoping we have

$$\sum_{n=0}^\infty \frac{\tan(a/2^n)}{2^n}=\lim_{n\to\infty}u_n-u_{-1}=\frac{1}{\tan a}- \frac{2}{\tan (2a)}=\tan a$$