Sum of Sequence of Products of Squares of 3 Consecutive Reciprocals

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Theorem

\(\ds \sum_{j \mathop = 1}^\infty \frac 1 {j^2 \paren {j + 1}^2 \paren {j + 2}^2}\) \(=\) \(\ds \frac 1 {1^2 \times 2^2 \times 3^2} + \frac 1 {2^2 \times 3^2 \times 4^2} + \frac 1 {3^2 \times 4^2 \times 5^2} + \frac 1 {4^2 \times 5^2 \times 6^2} + \cdots\)
\(\ds \) \(=\) \(\ds \frac {4 \pi^2 - 39} {16}\)


Proof 1

We use partial fraction expansion to expand $\dfrac 1 {j^2 \paren {j + 1}^2 \paren {j + 2}^2}$.

Let $\dfrac 1 {j^2 \paren {j + 1}^2 \paren {j + 2}^2} = \dfrac A j + \dfrac B {j^2} + \dfrac C {j + 1} + \dfrac D {\paren {j + 1}^2} + \dfrac E {j + 2} + \dfrac F {\paren {j + 2}^2}$.

This has solutions $A = - \dfrac 3 4, B = \dfrac 1 4, C = 0, D = 1, E = \dfrac 3 4, F = \dfrac 1 4$.

Thus:

\(\ds \sum_{j \mathop = 1}^\infty \frac 1 {j^2 \paren {j + 1}^2 \paren {j + 2}^2}\) \(=\) \(\ds \sum_{j \mathop = 1}^\infty \paren {-\frac 3 {4j} + \frac 1 {4 j^2} + \frac 1 {\paren {j + 1}^2} + \frac 3 {4 \paren {j + 2} } + \frac 1 {4 \paren {j + 2}^2} }\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^\infty \paren {\frac 1 {4 j^2} + \frac 1 {\paren {j + 1}^2} + \frac {3 j - 3 \paren {j + 2} } {4 j \paren {j + 2} } + \frac 1 {4 \paren {j + 2}^2} }\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^\infty \frac 1 {4 j^2} + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {j + 1}^2} + \sum_{j \mathop = 1}^\infty \frac 1 {4 \paren {j + 2}^2} - \sum_{j \mathop = 1}^\infty \frac 6 {4 j \paren {j + 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 4 \sum_{j \mathop = 1}^\infty \frac 1 {j^2} + \sum_{j \mathop = 2}^\infty \frac 1 {j^2} + \frac 1 4 \sum_{j \mathop = 3}^\infty \frac 1 {j^2} - \frac 3 2 \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 2} }\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds \frac 3 2 \sum_{j \mathop = 1}^\infty \frac 1 {j^2} - 1 - \frac 1 4 - \frac 1 {16} - \frac 3 2 \paren {\frac 3 4}\) Corollary to Sum of Sequence of Products of Consecutive Odd and Consecutive Even Reciprocals
\(\ds \) \(=\) \(\ds \frac 3 2 \paren {\frac {\pi^2} 6} - \frac {21} {16} - \frac 9 8\) Basel Problem
\(\ds \) \(=\) \(\ds \frac {4 \pi^2 - 39} {16}\)

$\blacksquare$


Proof 2

Note that we have:

\(\ds \frac 1 {2 j} - \frac 1 {j + 1} + \frac 1 {2 \paren {j + 2} }\) \(=\) \(\ds \frac {\paren {j + 1} \paren {j + 2} - 2 j \paren {j + 2} + j \paren {j + 1} } {2 j \paren {j + 1} \paren {j + 2} }\)
\(\ds \) \(=\) \(\ds \frac {j^2 + 3 j + 2 - 2 j^2 - 4 j + j^2 + j} {2 j \paren {j + 1} \paren {j + 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 {j \paren {j + 1} \paren {j + 2} }\)


Thus:

\(\ds \sum_{j \mathop = 1}^\infty \frac 1 {j^2 \paren {j + 1}^2 \paren {j + 2}^2}\) \(=\) \(\ds \sum_{j \mathop = 1}^\infty \paren {\frac 1 {2 j} - \frac 1 {j + 1} + \frac 1 {2 \paren {j + 2} } }^2\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^\infty \paren {\frac 1 {4 j^2} + \frac 1 {\paren {j + 1}^2} + \frac 1 {4 \paren {j + 2}^2} - \frac 2 {2 j \paren {j + 1} } - \frac 2 {2 \paren {j + 1} \paren {j + 2} } + \frac 2 {4 j \paren {j + 2} } }\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^\infty \frac 1 {4 j^2} + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {j + 1}^2} + \sum_{j \mathop = 1}^\infty \frac 1 {4 \paren {j + 2}^2} - \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 1} } - \sum_{j \mathop = 1}^\infty \frac 1 {\paren {j + 1} \paren {j + 2} } + \sum_{j \mathop = 1}^\infty \frac 1 {2 j \paren {j + 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 4 \sum_{j \mathop = 1}^\infty \frac 1 {j^2} + \sum_{j \mathop = 2}^\infty \frac 1 {j^2} + \frac 1 4 \sum_{j \mathop = 3}^\infty \frac 1 {j^2} - \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 1} } - \sum_{j \mathop = 2}^\infty \frac 1 {j \paren {j + 1} } + \frac 1 2 \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 2} }\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds \frac 3 2 \sum_{j \mathop = 1}^\infty \frac 1 {j^2} - 1 - \frac 1 4 - \frac 1 {16} - 2 \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 1} } + \frac 1 {1 \times 2} + \frac 1 2 \paren {\frac 3 4}\) Sum of Sequence of Products of Consecutive Odd and Consecutive Even Reciprocals/Corollary
\(\ds \) \(=\) \(\ds \frac 3 2 \paren {\frac {\pi^2} 6} - \frac {21} {16} - 2 \times 1 + \frac 1 2 + \frac 3 8\) Basel Problem, Corollary to Sum of Sequence of Products of Consecutive Reciprocals
\(\ds \) \(=\) \(\ds \frac {4 \pi^2 - 39} {16}\)

$\blacksquare$


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