Principle of Mathematical Induction/Warning/Example 2
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Example of Incorrect Use of Principle of Mathematical Induction
We are to prove that:
- $\dfrac 1 {1 \times 2} + \dfrac 1 {2 \times 3} + \dotsb + \dfrac 1 {\paren {n - 1} \times n} = \dfrac 3 2 - \dfrac 1 n$
For $n = 1$ we have:
- $\dfrac 3 2 - \dfrac 1 n = \dfrac 1 2 = \dfrac 1 {1 \times 2}$
Assuming true for $k$, we have:
\(\ds \dfrac 1 {1 \times 2} + \dfrac 1 {2 \times 3} + \dotsb + \dfrac 1 {\paren {n - 1} \times n} + \dfrac 1 {n \times \paren {n + 1} }\) | \(=\) | \(\ds \dfrac 3 2 - \frac 1 n + \dfrac 1 {n \paren {n + 1} }\) | by the induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 3 2 - \frac 1 n + \paren {\dfrac 1 n - \dfrac 1 {n + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 3 2 - \frac 1 {n + 1}\) |
But clearly this is wrong, because for $n = 6$:
- $\dfrac 1 2 + \dfrac 1 6 + \dfrac 1 {12} + \dfrac 1 {30} = \dfrac 5 6$
on the left hand side, but:
- $\dfrac 3 2 - \dfrac 1 6 = \dfrac 4 3$
on the right hand side.
Refutation
The supposed sequence of terms on the left hand side starts at $n = 2$.
It can be seen that it is meaningless (or has no terms for $n = 1$.
Hence in the first statement:
- $\dfrac 3 2 - \dfrac 1 n = \dfrac 1 2 = \dfrac 1 {1 \times 2}$
the term on the right hand side is the term for $n = 2$, making this equation invalid.
The correct result is Sum of Sequence of Products of Consecutive Reciprocals:
- $\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.1$: Mathematical Induction: Exercise $3$