Given ∑ n = 1 ∞ 1/ n 2 = π 2 /6 , find ∑ n = 0 ∞ 1/ ( 2 n + 1 ) 2 .

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Given $\sum_{n = 1}^{\infty} 1/ n^{2} = π^{2} /6$ , find $\sum_{n = 0}^{\infty} 1/ (2 n + 1)^{2}$ .

Answer:

Solution:

The even-index contribution is one quarter of the full sum, so the odd-index sum is $(1 - 1/4) π^{2} /6 = π^{2} /8$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: QB

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