The sum of the series frac{1}{2 times 3} + fr | vedclass

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

Question

(B) The given series is $S = \frac{1}{2 	imes 3} + \frac{1}{4 	imes 5} + \frac{1}{6 	imes 7} + \dots$ Using partial fractions,$\frac{1}{n(n+1)} = \

Vedclass · Accepted Answer

$\log(e/2)$

Vedclass · Answer

(B) The given series is $S = \frac{1}{2 	imes 3} + \frac{1}{4 	imes 5} + \frac{1}{6 	imes 7} + \dots$ Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Thus,$S = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{4} - \frac{1}{5}) + (\frac{1}{6} - \frac{1}{7}) + \dots$ We know the logarithmic series expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ For $x=1$,$\log_e(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots$ Rearranging this,$1 - \log_e(2) = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots$ This matches our series $S$. Therefore,$S = 1 - \log_e(2) = \log_e(e) - \log_e(2) = \log_e(e/2)$.

The sum of the series $\frac{1}{2 \times 3} + \frac{1}{4 \times 5} + \frac{1}{6 \times 7} + \dots = $

Similar Questions

$e^{\left( {x - \frac{1}{2}{(x - 1)}^2 + \frac{1}{3}{(x - 1)}^3 - \frac{1}{4}{(x - 1)}^4 + \dots} \right)}$ is equal to

If $|x| < 1$,then the coefficient of $x^5$ in the expansion of $(1 - x) \ln(1 - x)$ is

$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{3 \cdot 4 \cdot 5} + \frac{1}{5 \cdot 6 \cdot 7} + \dots \infty = $

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module