# Sum of Sequence of Products of Consecutive Odd Reciprocals

## 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 $\ds$ $=$ $\ds \frac 1 2 \paren {\frac 1 {2 \times 0 + 1} - \frac 1 {2 n + 3} }$ $\ds$ $=$ $\ds \frac 1 2 - \frac 1 {4 n + 6}$ $\ds$ $=$ $\ds \frac {\paren {2 n + 3} - 1} {4 n + 6}$ $\ds$ $=$ $\ds \frac {2 n + 2} {4 n + 6}$ $\ds$ $=$ $\ds \frac {n + 1} {2 n + 3}$

$\blacksquare$