WBJEE · Maths · Sequences and Series
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
- A \(\log _{e} 4\)
- B \(\log _{e} 2-1\)
- C \(1-\log _{e} 2\)
- D \(1-\log _{e} 4\)
Answer & Solution
Correct Answer
(D) \(1-\log _{e} 4\)
Step-by-step Solution
Detailed explanation
\(S_{1}=1+\frac{1}{2}+\frac{1}{2^{2}}+\infty=\frac{1}{1-\frac{1}{2}}=2\) \(S_{2}=2+2 \cdot \frac{2}{3}+2\left(\frac{2}{3}\right)^{2}+\ldots \infty=\frac{2}{1-\frac{2}{3}}=6\) \(S_{3}=3+3\left(\frac{3}{4}\right)+3\left(\frac{3}{4}\right)^{2}+\infty=\frac{3}{1-\frac{3}{4}}=12\)…
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