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

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(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)} = \

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$\log(e/2)$

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(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 = $

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