$\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 = $

  • A
    $\log_e x$
  • B
    $\log_e (1 + x)$
  • C
    $\log_e (1 - x)$
  • D
    $\log_e \frac{x}{1 + x}$

Explore More

Similar Questions

The value of the infinite series $\log _4 2 - \log _8 2 + \log _{16} 2 - \dots \infty$ is:

Difficult
View Solution

Evaluate the sum of the series: $\log_e \frac{4}{5} + \frac{1}{4} - \frac{1}{2} \left( \frac{1}{4} \right)^2 + \frac{1}{3} \left( \frac{1}{4} \right)^3 - \dots$

If $\alpha, \beta$ are the roots of the equation $x^2 - px + q = 0$,then $\log_e(1 + px + qx^2) = $

The value of the series $x \log _e a + \frac{x^3}{3!} (\log _e a)^3 + \frac{x^5}{5!} (\log _e a)^5 + \dots$ is

If $0 < y < 2^{1/3}$ and $x(y^3 - 1) = 1$,then $\frac{2}{x} + \frac{2}{3x^3} + \frac{2}{5x^5} + \dots$ is equal to:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo