Math  /  Calculus

Question8) Which of the following series is conditionally convergent? A) k=1(1)k1ln(k+1)\sum_{k=1}^{\infty}(-1)^{k} \frac{1}{\ln (k+1)} B) k=1(1)k1k2+1\sum_{k=1}^{\infty}(-1)^{k} \frac{1}{k^{2}+1} C) k=1(1)kk2k+1\sum_{k=1}^{\infty}(-1)^{k} \frac{k}{2 k+1} D) k=1(1)k(2k+1k+1)k\sum_{k=1}^{\infty}(-1)^{k}\left(\frac{2 k+1}{k+1}\right)^{k}

Studdy Solution
After evaluating all options, the only series that is conditionally convergent is Option A.
Thus, the correct answer is:
Option A: k=1(1)k1ln(k+1) \text{Option A: } \sum_{k=1}^{\infty}(-1)^{k} \frac{1}{\ln (k+1)}

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