The sum of an infinite geometric series whose first term  $\underset{x\to 0}{\lim }\sum _{K=1}^{2013}\frac{\{\frac{x}{\tan x}+2K\}}{2013}$ and whose common ratio is the value of  $\underset{x\to 0}{\lim }\frac{e^{{\tan }^{3}x}-e^{x^3}}{2\ln (1+x^3{\sin }^{2}x)}$  is ……

 $\underset{x\to 0}{\lim }\sum _{K=1}^{2013}\frac{\{\frac{x}{\tan x}+2K\}}{2013}$
 $=\underset{x\to 0}{\lim }\frac{\left\{\frac{x}{\tan x}\right\}\times 2013}{2013}$
 $=\underset{x\to 0}{\lim }\left\{\frac{x}{\tan x}\right\}[\because 0<\frac{x}{\tan x}<1]$
=1
   Ist term GP=1
Now, $\underset{x\to 0}{\lim }\frac{e^{{\tan }^{3}x}-e^{x^3}}{2\ln (1+x^3{\sin }^{2}x)}$
 = $\underset{x\to 0}{\lim }\frac{e^{{\tan }^{3}x-x^3}-1}{2\ln (1+x^3{\sin }^{2}x)}$
 $=\underset{x\to 0}{\lim }\frac{e^{x^3}(e^{{\tan }^{3x-x^3}}-1)({\tan }^{3}x-x^3)}{({\tan }^{3}x-x^3)}.\frac{1}{2\frac{\ln (1+x^3{\sin }^{2}x)}{x^3{\sin }^{2}x}{x}^3{\sin }^{2}x0}$
 $$\begin{gathered} =\underset{x\to 0}{\lim }\frac{1}{2}\left(\frac{{\tan }^{3}x-x^3}{x^3{\sin }^{2}x}\right) \\ =\underset{x\to 0}{\lim }\frac{1}{2}\left(\frac{\tan x-x}{x^3}\right)\left(\frac{{\tan }^{2}x+x^2+x\tan x}{x^2}\right)\left(\frac{x^2}{{\sin }^{2}x}\right) \\ =\frac{1}{2}.\frac{1}{3}\mathrm{.3.1}=\frac{1}{2} \end{gathered}$$
Common ration = $\frac{1}{2}$
 $\therefore S_{\infty }=\frac{1}{1-\frac{1}{2}}=\frac{1}{\left(\frac{1}{2}\right)}=2$