Sum of Sequence of Products of Consecutive Odd and Consecutive Even Reciprocals

Theorem

 $\ds \sum_{j \mathop = 1}^n \frac 1 {j \left({j + 2}\right)}$ $=$ $\ds \frac 1 {1 \times 3} + \frac 1 {2 \times 4} + \frac 1 {3 \times 5} + \frac 1 {4 \times 6} + \cdots + \frac 1 {n \left({n + 2}\right)}$ $\ds$ $=$ $\ds \frac 3 4 - \frac {2 n + 3} {2 \left({n + 1}\right) \left({n + 2}\right)}$

Corollary

 $\ds \sum_{j \mathop = 1}^\infty \frac 1 {j \left({j + 2}\right)}$ $=$ $\ds \frac 1 {1 \times 3} + \frac 1 {2 \times 4} + \frac 1 {3 \times 5} + \frac 1 {4 \times 6} + \cdots$ $\ds$ $=$ $\ds \frac 3 4$

Proof

We observe that:

 $\ds \frac 1 j - \frac 1 {j + 2}$ $=$ $\ds \frac {\left({j + 2}\right) - j} {j \left({j + 2}\right)}$ $\ds$ $=$ $\ds \frac 2 {\left({j \left({j + 2}\right)}\right)}$

Therefore:

 $\ds \sum_{j \mathop = 1}^n \frac 1 {j \left({j + 2}\right)}$ $=$ $\ds \frac 1 2 \sum_{j \mathop = 1}^n \left({\frac 1 j - \frac 1 {j + 2} }\right)$ $\ds$ $=$ $\ds \frac 1 2 \left({\left({\frac 1 1 - \frac 1 3}\right) + \left({\frac 1 2 - \frac 1 4}\right) + \left({\frac 1 3 - \frac 1 5}\right) + \cdots + \left({\frac 1 {n - 1} - \frac 1 {n + 1} }\right) + \left({\frac 1 n - \frac 1 {n + 2} }\right)}\right)$ $\ds$ $=$ $\ds \frac 1 2 \left({1 + \frac 1 2 - \frac 1 {n + 1} - \frac 1 {n + 2} }\right)$ Telescoping Series $\ds$ $=$ $\ds \frac 3 4 - \frac {\left({n + 2}\right) + \left({n + 1}\right)} {2 \left({n + 1}\right) \left({n + 2}\right)}$ $\ds$ $=$ $\ds \frac 3 4 - \frac {2 n + 3} {2 \left({n + 1}\right) \left({n + 2}\right)}$

$\blacksquare$