What is the value of: $$\sum_{k=0}^\infty \frac{1}{(4k+1)^2}?$$ I realised that $$\sum_{n=2,4,6,8,...} \frac{1}{n^2} + \sum_{n=1,3,5,7,...} \frac{1}{n^2} = \sum_{n \geq 1 } \frac{1}{n^2}$$ $$\sum_{n \geq 1 } \frac{1}{4n^2}+\sum_{n \geq 0 } \frac{1}{(2n+1)^2} = \sum_{n \geq 1 } \frac{1}{n^2}$$ $$\sum_{n \geq 0 } \frac{1}{(2n+1)^2} = \frac{3}{4}\frac{\pi^2}{6} \qquad \Rightarrow \qquad \sum_{n=1,3,5,7,...} \frac{1}{n^2} = \frac{1}{8} \cdot \pi^2$$ So $$\sum_{k=0}^\infty \frac{1}{(4k+1)^2} + \sum_{k=0}^\infty \frac{1}{(4k+3)^2} = \frac{\pi^2}{8}$$ But I cannot find the value of the second summation. Any suggestions?
2026-05-15 04:10:43.1778818243
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Question about finding the value of an infinite sum
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The sum can be expressed in terms of Catalan's constant. In the following video, the sum is computed in a step by step manner: https://www.youtube.com/watch?v=r2OJtsHNDZA.
Recall the definition of Catalan's Constant: $$K=\sum_{k=0}^{\infty} \frac 1 {(4k+1)^2}- \sum_{k=0}^{\infty} \frac 1 {(4k+3)^2}$$ Combining that with the equation you have already derived we immediately get $$\sum_{k=0}^{\infty} \frac 1 {(4k+1)^2}=\frac K2+\frac {\pi^2}{16}$$