The sum of the series frac{1}{1 times 2} - fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the series be $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n(n+1)}$.Using partial fractions, $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.

Vedclass · Accepted Answer

$2 \log_{e} 2 - 1$

Vedclass · Answer

(C) Let the series be $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n(n+1)}$.Using partial fractions, $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.Thus, $S = (\frac{1}{1} - \frac{1}{2}) - (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) - (\frac{1}{4} - \frac{1}{5}) + \dots$$S = 1 - \frac{1}{2} - \frac{1}{2} + \frac{1}{3} + \frac{1}{3} - \frac{1}{4} - \frac{1}{4} + \dots$$S = 1 - 2(\frac{1}{2}) + 2(\frac{1}{3}) - 2(\frac{1}{4}) + 2(\frac{1}{5}) - \dots$$S = 1 - 2[\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots]$We know the expansion $\log_{e}(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$For $x=1$, $\log_{e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$So, $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_{e} 2$.Substituting this back, $S = 1 - 2(1 - \log_{e} 2) = 1 - 2 + 2 \log_{e} 2 = 2 \log_{e} 2 - 1$.

The sum of the series $\frac{1}{1 \times 2} - \frac{1}{2 \times 3} + \frac{1}{3 \times 4} - \dots \infty$ is

Similar Questions

If $4\left[ {{x^2} + \frac{{{x^6}}}{3} + \frac{{{x^{10}}}}{5} + \dots} \right] = {y^2} + \frac{{{y^4}}}{2} + \frac{{{y^6}}}{3} + \dots$,then

If $y = x - \frac{x^2}{2!} + \frac{x^3}{3!} - \frac{x^4}{4!} + \dots$,then $x = $

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:

$\log_e [(1 + x)^{1 + x} (1 - x)^{1 - x}] = $

The expansion $\log_e(1 + x) = \sum\limits_{i = 1}^\infty \left[ \frac{(-1)^{i + 1}x^i}{i} \right]$ is defined for:

Vedclass Test Series

Exam Paper Generator

Online Exam Module