Sum of Sequence of Products of 3 Consecutive Reciprocals

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} } = \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} }$


Corollary

$\ds \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 1} \paren {j + 2} } = \frac 1 4$


Proof

We observe that:

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

Hence:

\(\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} }\) \(=\) \(\ds \frac 1 2 \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 2 {j + 1} + \frac 1 {j + 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\sum_{j \mathop = 1}^n \frac 1 j - \sum_{j \mathop = 1}^n \frac 2 {j + 1} + \sum_{j \mathop = 1}^n \frac 1 {j + 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\sum_{j \mathop = 1}^n \frac 1 j - \sum_{j \mathop = 2}^{n + 1} \frac 2 j + \sum_{j \mathop = 3}^{n + 2} \frac 1 j}\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac 1 1 + \frac 1 2 + \sum_{j \mathop = 3}^n \frac 1 j - \frac 2 2 - \frac 2 {n + 1} - \sum_{j \mathop = 3}^n \frac 2 j + \frac 1 {n + 1} + \frac 1 {n + 2} + \sum_{j \mathop = 3}^n \frac 1 j}\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac 1 1 + \frac 1 2 - \frac 2 2 - \frac 2 {n + 1} + \frac 1 {n + 1} + \frac 1 {n + 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac 1 2 - \frac 1 {n + 1} + \frac 1 {n + 2} }\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} \paren {n + 2} - 2 \paren {n + 2} + 2 \paren {n + 1} } {4 \paren {n + 1} \paren {n + 2} }\)
\(\ds \) \(=\) \(\ds \frac {n^2 + 3 n + 2 - 2 n - 4 + 2 n + 2} {4 \paren {n + 1} \paren {n + 2} }\)
\(\ds \) \(=\) \(\ds \frac {n^2 + 3 n} {4 \paren {n + 1} \paren {n + 2} }\)
\(\ds \) \(=\) \(\ds \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} }\)

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_of_Sequence_of_Products_of_3_Consecutive_Reciprocals&oldid=587279"