Infinite trigonometric summation Can the following summation be written in a finite number of terms:
$$\sum_{r=1}^{\infty}\frac{\tan(\theta/2^r)}{2^{r-1}\cos(\theta/2^{r-1})}$$
I tried to simplify the expression using trigonometric identities and then converting the infinite summation into a definite integral. I couldn't think of any way to get it into that since the variable $r$ occurs as an exponent.
So is there any other possible simplification for the expression? 

Let $a_n = \dfrac{\tan(\frac{x}{2^n})}{2^{n-1}\cos(x/2^{n-1})}$. We want to find $\sum_{n=1}^{\infty} a_n$.
Let's write out the first term: 
$$a_1 = \frac{\tan(x/2)}{\cos(x)}$$
and invoke the identity $\tan(\frac{x}{2}) = \csc(x)-\cot(x)$:
$$a_1 = \frac{\csc(x)-\cot(x)}{\cos(x)} \\ = \frac{1}{\sin(x)\cos(x)} - \frac{\cos(x)}{\sin(x)\cos(x)} \\ = \frac{2}{2\sin(x)\cos(x)} - \frac{1}{\sin(x)} \\ = \frac{2}{\sin(2x)} - \csc(x) \\ a_1 = 2\csc(2x) - \csc(x).$$
By performing the same procedure for the next couple terms ($a_2$ and $a_3$), we have the following:
$$a_2 = \csc(x) - \frac{1}{2}\csc(x/2) \\ 
a_3 = \frac{1}{2}\csc(x/2) - \frac{1}{4}\csc(x/4) \\ \vdots \\ \vdots \\ a_n = \frac{1}{2^{n-2}} \csc(x/2^{n-2}) - \frac{1}{2^{n-1}}\csc(x/2^{n-1}).$$

Now it's time to consider partial sums. Let $s_n = a_1 + a_2 + \cdots + a_n$. A pattern emerges quickly because of the telescoping nature of the terms:
$$s_1 = a_1 = 2\csc(2x) - \csc(x) \\ \ \\ \ \\ s_2 = s_1 + a_2  = [2\csc(2x) - \csc(x)] + [\csc(x) - \frac{1}{2}\csc(x/2)] \\ s_2 = 2\csc(2x) - \frac{1}{2}\csc(x/2) \\ \ \\ \ \\ s_3 = s_2 + a_3 = [2\csc(2x) - \frac{1}{2}\csc(x/2)] + [\frac{1}{2}\csc(x/2) - \frac{1}{4}\csc(x/4)] \\ s_3 = 2\csc(2x) - \frac{1}{4}\csc(x/4) \\ \vdots \\ \vdots \\ s_n = 2\csc(2x) - \frac{1}{2^{n-1}}\csc(x/2^{n-1}).$$

Now that we have an expression for the $n$-th partial sum, we can attempt considering the limiting case since $\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} s_n$. That is, 
$$\lim_{n \to \infty} s_n \\ 
= \lim_{n \to \infty} \left[ 2\csc(2x) - \frac{1}{2^{n-1}}\csc(x/2^{n-1}) \right] \\ 
= 2\csc(2x) - \lim_{n \to \infty} \left[\frac{1}{2^{n-1}}\csc(x/2^{n-1}) \right] \\
= 2\csc(2x) - \lim_{n \to \infty} \left[\frac{1}{2^{n-1}\sin(x/2^{n-1})}\right]
$$
Unfortunately, this limit is not defined. However, for $x\neq 0$, it can be shown that we're left with 
$$\lim_{n \to \infty} s_n = 2\csc(2x) - \frac{1}{x}.$$
So altogether, and replacing $n$ with $r$ and replacing $x$ with $\theta$, we have $$ \sum_{r=1}^{\infty} \frac{\tan(\frac{\theta}{2^r})}{2^{r-1}\cos(\theta/2^{r-1})} = 2\csc(2\theta) - \frac{1}{\theta}.$$
I'm not sure how to find out the restrictions on $x$ this late at night. Fro example, since $x=0$ gives zero for the sum but not this answer, perhaps someone else could take over at this point. However, I hope the derivation is of some use to you!
