If the sum of an infinite geometric series whose first term is  $\underset{x\to 0}{\lim }\sum _{k-1}^{2013}\frac{\{\frac{x}{\tan x}+2k\}}{2013}$  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 S then the value of  3S  is (Note : where $\left\{y\right\}$  denotes fractional pat of  y.)

$\underset{x\to 0}{\lim }\sum _{k-1}^{2013}\frac{\{\frac{x}{\tan x}+2k\}}{2013}=\underset{x\to 0}{\lim }\left\{\frac{x}{\tan x}\right\} =\underset{x\to 0}{\lim }(\frac{x}{\tan x}-\left[\frac{x}{\tan x}\right])=\underset{x\to 0}{\lim }\left(\frac{x}{\tan x}\right)=1$

$\therefore$1st term of G.P. = 1
common ratio = $\underset{x\to 0}{\lim }\frac{e^{x^3}(e^{{\tan }^{3}x-x^3}-1)({\tan }^{3}x-x^3)}{\frac{2\ln (1+x^3{\sin }^{2}x)}{x^3{\sin }^{2}x}.x^3{\sin }^{2}x.({\tan }^{3}x-x^3)}$
$$\begin{gathered} =\frac{1}{2}\underset{x\to 0}{\lim }\left(\frac{\tan x-x}{x^3}\right)\left(\frac{{\tan }^{2}x+x^2+x\tan x}{x^2}\right)=\frac{1}{2}.\frac{1}{3}\times 3=\frac{1}{2} \\ \therefore S=\frac{1}{1-\left(\frac{1}{2}\right)}=2 \end{gathered}$$