Sum of Sequence of Products of Squares of Consecutive Odd Reciprocals

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Theorem

\(\ds \sum_{j \mathop = 0}^\infty \frac 1 {\left({2 j + 1}\right)^2 \left({2 j + 3}\right)^2}\) \(=\) \(\ds \frac 1 {1^2 \times 3^2} + \frac 1 {3^2 \times 5^2} + \frac 1 {5^2 \times 7^2} + \frac 1 {7^2 \times 9^2} + \cdots\)
\(\ds \) \(=\) \(\ds \frac {\pi^2 - 8} {16}\)


Proof

Note that we have:

$\dfrac 1 {2 j + 1} - \dfrac 1 {2 j + 3} = \dfrac 2 {\paren {2 j + 1} \paren {2 j + 3} }$

So:

\(\ds \sum_{j \mathop = 0}^\infty \frac 1 {\paren {2 j + 1}^2 \paren {2 j + 3}^2}\) \(=\) \(\ds \frac 1 4 \sum_{j \mathop = 0}^\infty \paren {\frac 1 {2 j + 1} - \frac 1 {2 j + 3} }^2\)
\(\ds \) \(=\) \(\ds \frac 1 4 \sum_{j \mathop = 0}^\infty \paren {\frac 1 {\paren {2 j + 1}^2} - \frac 2 {\paren {2 j + 1} \paren {2 j + 3} } + \frac 1 {\paren {2 j + 3}^2} }\) Square of Sum
\(\ds \) \(=\) \(\ds \frac 1 4 \sum_{j \mathop = 0}^\infty \frac 1 {\paren {2 j + 1}^2} - \frac 1 2 \sum_{j \mathop = 0}^\infty \frac 1 {\paren {2 j + 1} \paren {2 j + 3} } + \frac 1 4 \paren {\sum_{j \mathop = 0}^\infty \frac 1 {\paren {2 j + 1}^2} - 1}\)
\(\ds \) \(=\) \(\ds \frac {\pi^2} {32} - \frac 1 4 + \frac {\pi^2} {32} - \frac 1 4\) Sum of Reciprocals of Squares of Odd Integers, Sum of Sequence of Products of Consecutive Odd Reciprocals
\(\ds \) \(=\) \(\ds \frac {\pi^2} {16} - \frac 1 2\)
\(\ds \) \(=\) \(\ds \frac {\pi^2 - 8} {16}\)

$\blacksquare$


Sources

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