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

$(0.5) - \frac{(0.5)^2}{2} + \frac{(0.5)^3}{3} - \frac{(0.5)^4}{4} + \dots$

Evaluate: $\log _e(x + 1) - \log _e(x - 1) = $

Let $\alpha$ and $\beta$ be the roots of $5x^2 - 3x - 1 = 0$. Then the expression $\left[ (\alpha + \beta)x - \left( \frac{\alpha^2 + \beta^2}{2} \right)x^2 + \left( \frac{\alpha^3 + \beta^3}{3} \right)x^3 - \dots \right]$ is equal to:

$\frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5} + \dots \infty = $

Evaluate: $\log_e \sqrt{\frac{1+x}{1-x}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module