Evalúe $\sum_{r=1}^{n}\frac{\tan\left(\frac{x}{2^r}\right)}{2^{r-1}\cos\left(\frac{x}{2^{r-1}}\right)}$

Question

Evalúe: $$\sum_{r=1}^{n}\frac{\tan\left(\frac{x}{2^r}\right)}{2^{r-1}\cos\left(\frac{x}{2^{r-1}}\right)}$$

Mi intento:

Sea $$S=\sum_{r=1}^{n}\frac{\tan\left(\frac{x}{2^r}\right)}{2^{r-1}\cos\left(\frac{x}{2^{r-1}}\right)}$$ $$=\sum_{r=1}^{n}\frac{1}{2^{r-1}}\left(\tan\left(\frac{x}{2^{r-1}}\right)-\tan\left(\frac{x}{2^{r}}\right)\right)$$

¿Puedo utilizar la identidad trigonométrica

$\cot A-\tan A=-2\cot 2A$

o hay algún otro método

Answer 1

Nota

\begin{align} \frac{\tan\frac{x}{2^r}}{2^{r-1}\cos\frac{x}{2^{r-1}}} &= \frac{\sin\frac{x}{2^r}}{2^{r-1} \cos\frac{x}{2^{r}}\cos\frac{x}{2^{r-1}}} = \frac{\sin^2\frac{x}{2^r}}{2^{r-1} \sin\frac{x}{2^{r}}\cos\frac{x}{2^{r}}\cos\frac{x}{2^{r-1}}}\\ &= \frac{1-\cos\frac{x}{2^{r-1}}}{2^{r-1} \sin\frac{x}{2^{r-1}}\cos\frac{x}{2^{r-1}}} = \frac{1}{2^{r-2} \sin\frac{x}{2^{r-2}}} - \frac{1}{2^{r-1} \sin\frac{x}{2^{r-1}}} \end{align}

Así, $$\sum_{r=1}^{n}\frac{\tan\frac{x}{2^r}}{2^{r-1}\cos\frac{x}{2^{r-1}}}=\frac2{\sin2{x}} - \frac{1}{2^{n-1} \sin\frac{x}{2^{n-1}}} $$