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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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$1-\log _{e} 4$
\(S_{1}=1+\frac{1}{2}+\frac{1}{2^{2}}+\infty=\frac{1}{1-\frac{1}{2}}=2\)
\(S_{2}=2+2 \cdot \frac{2}{3}+2\left(\frac{2}{3}\right)^{2}+\ldots \infty=\frac{2}{1-\frac{2}{3}}=6\)
\(S_{3}=3+3\left(\frac{3}{4}\right)+3\left(\frac{3}{4}\right)^{2}+\infty=\frac{3}{1-\frac{3}{4}}=12\)
\(S_{4}=4+4\left(\frac{4}{5}\right)+4\left(\frac{4}{5}\right)^{2}+\infty=\frac{4}{1-\frac{4}{5}}=20\)
\(\therefore \sum_{k=1}^{\infty} \frac{(-1)^{k}}{S_{k}}=-\frac{1}{S_{1}}+\frac{1}{S_{2}}-\frac{1}{S_{3}}+\frac{1}{S_{4}}-\infty\)
\(=-\frac{1}{2}+\frac{1}{6}-\frac{1}{12}+\frac{1}{20}-\infty\)
\(=-\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}-\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}-\infty\)
\(=-\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)-\ldots \infty\)
\(=-1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\infty\right)\)
\(=-1-2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots \infty\right)+2 \infty\)
\(=-1-2 \log _{e} 2+2=1-\log _{e} 4\)
\(S_{2}=2+2 \cdot \frac{2}{3}+2\left(\frac{2}{3}\right)^{2}+\ldots \infty=\frac{2}{1-\frac{2}{3}}=6\)
\(S_{3}=3+3\left(\frac{3}{4}\right)+3\left(\frac{3}{4}\right)^{2}+\infty=\frac{3}{1-\frac{3}{4}}=12\)
\(S_{4}=4+4\left(\frac{4}{5}\right)+4\left(\frac{4}{5}\right)^{2}+\infty=\frac{4}{1-\frac{4}{5}}=20\)
\(\therefore \sum_{k=1}^{\infty} \frac{(-1)^{k}}{S_{k}}=-\frac{1}{S_{1}}+\frac{1}{S_{2}}-\frac{1}{S_{3}}+\frac{1}{S_{4}}-\infty\)
\(=-\frac{1}{2}+\frac{1}{6}-\frac{1}{12}+\frac{1}{20}-\infty\)
\(=-\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}-\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}-\infty\)
\(=-\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)-\ldots \infty\)
\(=-1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\infty\right)\)
\(=-1-2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots \infty\right)+2 \infty\)
\(=-1-2 \log _{e} 2+2=1-\log _{e} 4\)
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