The sum of the infinite series frac{1}{1 time | vedclass

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(D) The $n^{th}$ term of the series is $T_n = \frac{(-1)^{n+1}}{n(n+1)}$.Using partial fractions,$T_n = (-1)^{n+1} \left( \frac{1}{n} - \frac{1}{n+1}

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$\log_e \left( \frac{4}{e} \right)$

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(D) The $n^{th}$ term of the series is $T_n = \frac{(-1)^{n+1}}{n(n+1)}$.Using partial fractions,$T_n = (-1)^{n+1} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.The sum $S = \sum_{n=1}^{\infty} (-1)^{n+1} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.Expanding the terms:$S = \left( 1 - \frac{1}{2} \right) - \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) - \dots$$S = 1 - 2 \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots \right)$.We know the expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$. For $x=1$,$\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$.Thus,$\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_e 2$.Substituting this back: $S = 1 - 2(1 - \log_e 2) = 1 - 2 + 2 \log_e 2 = 2 \log_e 2 - 1 = \log_e 4 - \log_e e = \log_e \left( \frac{4}{e} \right)$.

The sum of the infinite series $\frac{1}{1 \times 2} - \frac{1}{2 \times 3} + \frac{1}{3 \times 4} - \dots \infty$ is equal to:

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