Sum of Sequence of Products of Consecutive Reciprocals/Proof 3
Jump to navigation
Jump to search
Theorem
- $\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$
Proof
Observe that:
\(\ds \int_j^{j + 1} {\dfrac {\rd x} {x^2} }\) | \(=\) | \(\ds \intlimits {\dfrac {-1} x} j {j + 1}\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 j - \dfrac 1 {j + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {j \paren {j + 1} }\) |
Therefore:
\(\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} }\) | \(=\) | \(\ds \sum_{j \mathop = 1}^n \int_j^{j + 1} {\dfrac {\rd x} {x^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_1^{n + 1} {\dfrac {\rd x} {x^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \intlimits {\dfrac {-1} x} 1 {n + 1}\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac n {n + 1}\) |
$\blacksquare$