$\frac{x - 1}{x + 1} + \frac{1}{2} \cdot \frac{x^2 - 1}{(x + 1)^2} + \frac{1}{3} \cdot \frac{x^3 - 1}{(x + 1)^3} + \dots \infty = $

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

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

$\cosh^{-1} 2 = $

If $0 < a, b < 1$ and $\tan^{-1} a + \tan^{-1} b = \frac{\pi}{4}$,then the value of $(a+b) - \left(\frac{a^2+b^2}{2}\right) + \left(\frac{a^3+b^3}{3}\right) - \left(\frac{a^4+b^4}{4}\right) + \dots$ is ..... .

If $x, y, z$ are three consecutive positive integers,then $\frac{1}{2}\log_e x + \frac{1}{2}\log_e z + \frac{1}{2xz + 1} + \frac{1}{3}\left( \frac{1}{2xz + 1} \right)^3 + \dots = $