Sum of Sequence of Products of Consecutive Reciprocals/Proof 3

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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
\(\ds \) \(=\) \(\ds \frac n {n + 1}\)

$\blacksquare$