A topic related to infinity♾️.

What does \( 0.9999… \) (with an infinite number of \(9\)s after the decimal point) equal?

The answer is \( 0.9999… = 1 \). The proof can be easily shown by using an infinite geometric series.

Let the first term be \( a = 9/10 \), the common ratio be \( r = 1/10 \), and the sum of the first \( n+1 \) terms be \( S_n \). Then

\[ S_n = a + ra + r^2a + r^3a + … + r^n a, \]

\[ rS_n = ra + r^2a + r^3a + r^4a + … + r^{n+1}a \]

and by subtracting each side of the two equations, we obtain \( (1-r)S_n = a-r^{n+1}a \).

Since \( 0 < r < 1\), if \(n \to \infty \), then \[ \lim_{n \to \infty} S_n= \frac{a}{1-r}. \]

Therefore \[ \frac{9/10}{1-1/10} = 1. \]