Let S_{k} be the sum of an infinite GP series | vedclass

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(D) The sum of an infinite $GP$ with first term $a=k$ and common ratio $r=\frac{k}{k+1}$ is given by $S_{k} = \frac{a}{1-r} = \frac{k}{1-\frac{k}{k+1}

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$1-\log _{e} 4$

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(D) The sum of an infinite $GP$ with first term $a=k$ and common ratio $r=\frac{k}{k+1}$ is given by $S_{k} = \frac{a}{1-r} = \frac{k}{1-\frac{k}{k+1}} = \frac{k}{\frac{1}{k+1}} = k(k+1)$.We need to calculate $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{S_{k}} = \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k(k+1)}$.Using partial fractions,$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.So,the sum is $\sum_{k=1}^{\infty} (-1)^{k} \left( \frac{1}{k} - \frac{1}{k+1} \right) = -(\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$$= -1 + \frac{1}{2} + \frac{1}{2} - \frac{1}{3} - \frac{1}{3} + \frac{1}{4} + \frac{1}{4} - \dots$$= -1 + 2 \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots \right)$.Recall 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,the sum is $-1 + 2(1 - \log_{e} 2) = -1 + 2 - 2 \log_{e} 2 = 1 - 2 \log_{e} 2 = 1 - \log_{e} 4$.

Let $S_{k}$ be the sum of an infinite $GP$ series whose first term is $k$ and common ratio is $\frac{k}{k+1}$ $(k>0)$. Then,the value of $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{S_{k}}$ is equal to

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