Sum of Sequence of Products of Consecutive Odd Reciprocals
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Theorem
| \(\ds \sum_{j \mathop = 0}^n \frac 1 {\paren {2 j + 1} \paren {2 j + 3} }\) | \(=\) | \(\ds \frac 1 {1 \times 3} + \frac 1 {3 \times 5} + \frac 1 {5 \times 7} + \frac 1 {7 \times 9} + \cdots + \frac 1 {\paren {2 n + 1} \paren {2 n + 3} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {n + 1} {2 n + 3}\) |
Corollary
| \(\ds \sum_{j \mathop = 0}^\infty \frac 1 {\paren {2 j + 1} \paren {2 j + 3} }\) | \(=\) | \(\ds \frac 1 {1 \times 3} + \frac 1 {3 \times 5} + \frac 1 {5 \times 7} + \frac 1 {7 \times 9} + \cdots\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2\) |
Proof
We observe that:
| \(\ds \frac 1 {2 j + 1} - \frac 1 {2 j + 3}\) | \(=\) | \(\ds \frac {\paren {2 j + 3} - \paren{2 j + 1} } {\paren {2 j + 1} \paren {2 j + 3} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 2 {\paren {2 j + 1} \paren {2 j + 3} }\) |
and that $\ds \sum_{j \mathop = 0}^n \paren {\frac 1 {2 j + 1} - \frac 1 {2 j + 3} }$ is a telescoping series.
Therefore:
| \(\ds \sum_{j \mathop = 0}^n \frac 1 {\paren {2 j + 1} \paren {2 j + 3} }\) | \(=\) | \(\ds \frac 1 2 \sum_{j \mathop = 1}^n \paren {\frac 1 {2 j + 1} - \frac 1 {2 j + 3} }\) | Telescoping Series: Example 1 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {\frac 1 {2 \times 0 + 1} - \frac 1 {2 n + 3} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 - \frac 1 {4 n + 6}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {2 n + 3} - 1} {4 n + 6}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 n + 2} {4 n + 6}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {n + 1} {2 n + 3}\) |
$\blacksquare$