Calculation of Trigonometric Limit with Summation. If $\displaystyle f(x)=\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$ Then  value of $f(x)$ is 
$\bf{My\; Try::}$ Let $\displaystyle \left(\frac{x}{2^{r+1}}\right)=y$. So Trigonometric Expression is
$\displaystyle \frac{\tan y\cdot \left(1+\tan^2 y\right)}{\left(1-\tan^2 y\right)}=\frac{\tan y}{\cos 2y}=\frac{\sin y}{\cos 2y \cdot \cos y} = \frac{1}{2}\left\{\frac{\sin (2y-y)}{\cos 2y \cdot \cos y}\right\} = \frac{1}{2}\left\{\tan (2y)-\tan (y)\right\}$
Now How Can I solve after that
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$\displaystyle 2f(x)=2\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$
$\displaystyle=\sum_{r=0}^{n}\left\{\tan \frac x{2^r}-\tan \frac x{2^{r+1}}\right\}$ which is telescopic 
