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Normally when I keep try multiple ways to solve a problem, I get an idea of where to start, and eventually can solve it. But it hasn't been the way for this question and I've been stuck for hours.

I'm trying to determine the sum of the series $\sum_{n=1}^\infty{\frac{1}{2^n}\tan(\frac{x}{2^n})}$;

Yet, this isn't a geometric series, a telescoping series, and when I tried to expand it in the Fourier series in an attempt to reduce some of the factors, that didn't work either.

My professor seemed to give the hint: $\tan(x/2)$ = $\cot(x/2)$ - $2 $$\cot(x)$;

Yet this too is not really helping me because when I substitute this trig equation into the series, I get more of a mess as $\cot(x/2^n)$ is just as hard to solve as $\tan(x/2^n)$. So I thought this was a hint to help simplify the sum after I did most of the work.

We also haven't touched on complex numbers, so I don't think that's a solution to this as I've seen in other questions.

I would really appreciate a hint where to start without someone completely telling me how to solve this problem.

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Using the hint of your prof we get

$$\frac1{2^n}\tan\left(\frac x{2^n}\right)=\frac1{2^n}\cot\left(\frac x{2^n}\right)-\frac1{2^{n-1}}\cot\left(\frac x{2^{n-1}}\right)=u_n-u_{n-1}$$ and then telescope.

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To evaluate any infinite series, you need to first look at the partial sum.

Use induction and the identity given by your Professor to prove that $$\sum_{n=1}^N \dfrac1{2^n} \tan\left(\dfrac{x}{2^n}\right) = \dfrac1{2^N} \cot\left(\dfrac{x}{2^N}\right)-\cot(x)$$

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