Sum of Sequence of Products of Consecutive Reciprocals
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Theorem
- $\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$
Corollary
- $\ds \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = 1$
Proof 1
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
- $\ds \forall n \ge 1: \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$
Basis for the Induction
$\map P 1$ is true, as this just says $\dfrac 1 2 = \dfrac 1 2$.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\ds \sum_{j \mathop = 1}^k \frac 1 {j \paren {j + 1} } = \frac k {k + 1}$
Then we need to show:
- $\ds \sum_{j \mathop = 1}^{k + 1} \frac 1 {j \paren {j + 1} } = \frac {k + 1} {k + 2}$
Induction Step
This is our induction step:
\(\ds \sum_{j \mathop = 1}^{k + 1} \frac 1 {j \left({j + 1}\right)}\) | \(=\) | \(\ds \sum_{j \mathop = 1}^k \frac 1 {j \paren {j + 1} } + \frac 1 {\paren {k + 1} \paren {k + 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac k {k + 1} + \frac 1 {\paren {k + 1} \paren {k + 2} }\) | Induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {k \paren {k + 2} + 1} {\paren {k + 1} \paren {k + 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {k^2 + 2 k + 1} {\paren {k + 1} \paren {k + 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {k + 1}^2} {\paren {k + 1} \paren {k + 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {k + 1} {k + 2}\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \ge 1: \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$
$\blacksquare$
Proof 2
We can observe that:
- $\dfrac 1 {j \paren {j + 1} } = \dfrac 1 j - \dfrac 1 {j + 1}$
and that $\ds \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 1 {j + 1} }$ is a telescoping series.
Therefore:
\(\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} }\) | \(=\) | \(\ds \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 1 {j + 1} }\) | Telescoping Series: Example 1 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 - \frac 1 {n + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac n {n + 1}\) |
$\blacksquare$
Proof 3
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac n {n + 1}\) |
$\blacksquare$
Sources
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {1-1}$ Principle of Mathematical Induction: Exercise $6$
- 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.6$: Mathematical Induction: Problem Set $\text{A}.6$: $39$