The sum $\sum\limits_{n = 1}^\infty {{\cot }^{ - 1}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ is equal to

  • A
    $\frac{3\pi}{4} + \cot^{-1} 2$
  • B
    $\frac{\pi}{2} + \cot^{-1} 3$
  • C
    $\pi$
  • D
    $\frac{\pi}{2} + \tan^{-1} 2$

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Let $E_1 = \{x \in R : x \neq 1 \text{ and } \frac{x}{x-1} > 0\}$ and $E_2 = \{x \in E_1 : \sin^{-1}(\log_e(\frac{x}{x-1})) \text{ is a real number}\}$. (Here,the inverse trigonometric function $\sin^{-1} x$ assumes values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$). Let $f : E_1 \rightarrow R$ be the function defined by $f(x) = \log_e(\frac{x}{x-1})$ and $g : E_2 \rightarrow R$ be the function defined by $g(x) = \sin^{-1}(\log_e(\frac{x}{x-1}))$. Match the items in $LIST I$ with $LIST II$.
$LIST I$ $LIST II$
$P$. The range of $f$ is $1$. $(-\infty, \frac{1}{1-e}] \cup [\frac{e}{e-1}, \infty)$
$Q$. The range of $g$ contains $2$. $(0, 1)$
$R$. The domain of $f$ contains $3$. $[-\frac{1}{2}, \frac{1}{2}]$
$S$. The domain of $g$ is $4$. $(-\infty, 0) \cup (0, \infty)$
$5$. $(-\infty, \frac{e}{e-1}]$
$6$. $(-\infty, 0) \cup (\frac{1}{2}, \frac{e}{e-1}]$

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