# Sum of Sequence of Products of Squares of Consecutive Odd Reciprocals

## 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 $\ds$ $=$ $\ds \frac {\pi^2} {16} - \frac 1 2$ $\ds$ $=$ $\ds \frac {\pi^2 - 8} {16}$

$\blacksquare$