Limit involving trigonometric series summation

Question

How should I go about this one?

$\large{L = \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{2^k} \tan\left(\dfrac{\pi}{3\cdot2^{k+1}}\right)}$

Hints please!

Answer 1

Let $\displaystyle x= \frac{\pi}{6}\;,$ Then $\displaystyle L = \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{2^k}\tan\left(\frac{x}{2^k}\right)$

Now using $$\displaystyle \bullet \; \cot \left(\frac{x}{2}\right)-\tan \left(\frac{x}{2}\right) = 2\tan (x)\Rightarrow \tan \left(\frac{x}{2}\right)=\cot \left(\frac{x}{2}\right)-2\cot x$$

So $$\displaystyle \bullet \; \frac{1}{2}\tan\left(\frac{x}{2}\right) = \frac{1}{2}\cot \left(\frac{x}{2}\right)-\cot x$$

So $$\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{2^k}\tan\left(\frac{x}{2^k}\right) = \lim_{n\rightarrow \infty}\left[\frac{1}{2^n}\cot\left(\frac{x}{2^n}\right)-\cot x\right] = \frac{1}{x}-\cot x$$

So we get $$\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{2^k}\tan\left(\frac{x}{2^k}\right) = \frac{6}{\pi}-\sqrt{3}$$