A simple proof for \(0.999\ldots=1\).
Published on 28-05-2025
We have, \[0.999\ldots=\sum_{n=1}^{\infty} \frac{9}{10^n}=9(\frac{1}{1-\frac{1}{10}}-1)=9(\frac{10}{9}-1)=1.\]
We have, \[0.999\ldots=\sum_{n=1}^{\infty} \frac{9}{10^n}=9(\frac{1}{1-\frac{1}{10}}-1)=9(\frac{10}{9}-1)=1.\]