If $\tanh ^{-1} x = a \log \left(\frac{1+x}{1-x}\right)$, $|x| < 1$, then $a$ is equal to

For $|x| < 1$,the coefficient of $x^3$ in the expansion of $\log(1+x+x^2)$ in ascending powers of $x$ is:

If $0 < x < 1$ and $y = \frac{1}{2} x^{2} + \frac{2}{3} x^{3} + \frac{3}{4} x^{4} + \dots$,then the value of $e^{1+y}$ at $x = \frac{1}{2}$ is:

$1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \dots = $

$\frac{1}{2} - \frac{1}{2 \cdot 2^2} + \frac{1}{3 \cdot 2^3} - \frac{1}{4 \cdot 2^4} + \ldots$ is equal to

If $1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\frac{49-20 \sqrt{6}}{180}+\ldots$ upto $\infty = 2\left(\sqrt{\frac{b}{a}}+1\right) \log _e\left(\frac{a}{b}\right)$,where $a$ and $b$ are integers with $\operatorname{gcd}(a, b)=1$,then $11 a+18 b$ is equal to ...............