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Alternating p-series Given that , show that
Chapter 10, Problem 49E(choose chapter or problem)
QUESTION:
Alternating p-series Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\), show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}}=\frac{\pi^{2}}{12}\). (Assume the result of Exercise 53.)
Questions & Answers
QUESTION:
Alternating p-series Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\), show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}}=\frac{\pi^{2}}{12}\). (Assume the result of Exercise 53.)
ANSWER:Solution:-Step1Given that Step2show that Step