0.999...=1/Proof 1
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Theorem
- $0.999 \ldots = 1$
Proof
By Sum of Infinite Geometric Sequence:
- $0.999 \ldots = \dfrac a {1 - r}$
where $a = \dfrac 9 {10}$ and $r = \dfrac 1 {10}$.
Since our ratio is less than $1$, then we know that $\ds \sum_{n \mathop = 0}^\infty \frac 9 {10} \paren {\frac 1 {10} }^n$ must converge to:
- $\dfrac a {1 - r} = \dfrac {\frac 9 {10} } {1 - \frac 1 {10} } = \dfrac {\frac 9 {10} } {\frac 9 {10} } = 1$
$\blacksquare$