If S = frac{1}{1 times 2} - frac{1}{2 times 3 | vedclass

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(B) Given the series $S = \frac{1}{1 	imes 2} - \frac{1}{2 	imes 3} + \frac{1}{3 	imes 4} - \frac{1}{4 	imes 5} + \dots + \infty$. Using partial f

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$\frac{4}{e}$

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(B) Given the series $S = \frac{1}{1 	imes 2} - \frac{1}{2 	imes 3} + \frac{1}{3 	imes 4} - \frac{1}{4 	imes 5} + \dots + \infty$. Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. So,$S = (1 - \frac{1}{2}) - (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) - (\frac{1}{4} - \frac{1}{5}) + \dots$. $S = 1 - 2(\frac{1}{2}) + 2(\frac{1}{3}) - 2(\frac{1}{4}) + 2(\frac{1}{5}) - \dots$. $S = 1 - 2(\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots)$. We know that $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$. For $x=1$,$\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$,which implies $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_e 2$. Substituting this into $S$: $S = 1 - 2(1 - \log_e 2) = 1 - 2 + 2 \log_e 2 = 2 \log_e 2 - 1 = \log_e 4 - \log_e e = \log_e(\frac{4}{e})$. Therefore,$e^S = e^{\log_e(4/e)} = \frac{4}{e}$.

If $S = \frac{1}{1 \times 2} - \frac{1}{2 \times 3} + \frac{1}{3 \times 4} - \frac{1}{4 \times 5} + \dots + \infty$,then $e^S = $

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