frac{1}{1 cdot 2} - frac{1}{2 cdot 3} + frac{ | vedclass

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

Question

(A) We know the logarithmic series expansions:${\log_e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}

Vedclass · Accepted Answer

${\log_e} \frac{4}{e}$

Vedclass · Answer

(A) We know the logarithmic series expansions:${\log_e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$However,the given series is $S = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)}$.Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.So,$S = \sum_{n=1}^{\infty} (-1)^{n-1} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.$S = \left( 1 - \frac{1}{2} \right) - \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) - \left( \frac{1}{4} - \frac{1}{5} \right) + \dots$$S = 1 - 2\left( \frac{1}{2} \right) + 2\left( \frac{1}{3} \right) - 2\left( \frac{1}{4} \right) + \dots$$S = 1 - 2 \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots \right)$.Since ${\log_e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$,we have $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - {\log_e} 2$.Substituting this into $S$:$S = 1 - 2(1 - {\log_e} 2) = 1 - 2 + 2{\log_e} 2 = 2{\log_e} 2 - 1$.$S = {\log_e} 4 - {\log_e} e = {\log_e} \left( \frac{4}{e} \right)$.

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

Similar Questions

$\log _4 2 - \log _8 2 + \log _{16} 2 - \ldots$ is equal to

If $y = 2x^2 - 1$,then $\left[ \frac{1}{y} + \frac{1}{3y^3} + \frac{1}{5y^5} + \dots \right]$ is equal to

$\frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + \dots \infty = $

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module