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

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Question

(C) Let the series be $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n(n+1)}$.Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.T

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

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