Finding sum of a GP-tangent series.

Question

The question is to find the limit of the following sum when number of terms tend to infinity.$$\sum_{r=1}^\infty\frac{1}{2^r}\tan\left(\frac{\pi}{2^{r+1}}\right)$$I do not have any idea how to handel such series.It seems to be a combination of a GP and a trigonometric series.
I tried converting into secant and creating the difference($V_n$ method like) but no result.
If someone can give me a general formula for n terms it will be great.Sum of infinte terms will also do.
Please help.Thanks.

Answer 1

First look at the partial sum

$$\sum_{r=1}^N\frac{1}{2^r}\tan\left(\frac{\pi}{2^{r+1}}\right)$$

and replace $\tan(x)=\cot(x)-2\cot(2x)$ such that

$$\sum_{r=1}^N\left[\frac{1}{2^r}\cot\left(\frac{\pi}{2^{r+1}}\right)-\frac{1}{2^{r-1}}\cot\left(\frac{\pi}{2^{r}}\right)\right]$$

This sum telescopes and gives

$$\frac{1}{2^N}\cot\left(\frac{\pi}{2^{N+1}}\right)-\cot\left(\frac{\pi}{2}\right)$$

Remember that $\cot(\pi/2)=0$ we are therefore left with

$$\frac{1}{2^N}\cot\left(\frac{\pi}{2^{N+1}}\right)$$

Taking the limit $N\to+\infty$ we obtain

$$\lim_{N\to+\infty}\frac{1}{2^N}\cot\left(\frac{\pi}{2^{N+1}}\right)=\lim_{N\to+\infty}\frac{\cos\left(\frac{\pi}{2^{N+1}}\right)}{\frac{\sin\left(\frac{\pi}{2^{N+1}}\right)}{\frac{1}{2^N}}}=\frac{2}{\pi}\lim_{N\to+\infty}\frac{\cos\left(\frac{\pi}{2^{N+1}}\right)}{\frac{\sin\left(\frac{\pi}{2^{N+1}}\right)}{\frac{\pi}{2^{N+1}}}}=\frac{2}{\pi} \; .$$

Answer 2

An alternative solution. From $$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2 \pi^2}\right) \tag{1}$$ we have (by applying $\frac{d}{dx}\log(\cdot)$) $$ \tan(x)=\sum_{n\geq 0}\frac{8x}{(2n+1)^2\pi^2 - 4x^2}=-\sum_{n\geq 0}\left(\frac{1}{x-\frac{(2n+1)\pi}{2}}+\frac{1}{x+\frac{(2n+1)\pi}{2}}\right) \tag{2}$$

$$\begin{eqnarray*} \frac{1}{2^r}\tan\left(\frac{\pi}{2^{r+1}}\right)&=&\frac{2}{\pi}\sum_{n\geq 0}\left(\frac{1}{(2n+1)2^r-1}-\frac{1}{(2n+1)2^r+1}\right)\end{eqnarray*} \tag{3}$$ where every even number in $\mathbb{N}^*$ can be represented in a unique way as $(2n+1)2^r$.
It follows that by summing both sides of $(3)$ over $r\geq 1$ we have $$\sum_{r\geq 1}\frac{1}{2^r}\tan\left(\frac{\pi}{2^{r+1}}\right)=\frac{2}{\pi}\sum_{m\geq 1}\left(\frac{1}{2m-1}-\frac{1}{2m+1}\right)=\color{red}{\frac{2}{\pi}}\tag{4}$$ since the middle sum is a telescopic series.

A simple geometric interpretation is a quarter-circle can be decomposed as depicted.