0.99999 and it goes on forever

......I've lost count the number of times I answered this same question. Anyway...

$0.999999\ldots\\ =0.9 + 0.09 + 0.009 + \ldots$

This is a sum of a geometric progression to infinity, where

$a=0.9 \quad\quad r=\dfrac{0.09}{0.9} = 0.1$

Therefore,

$0.999999\ldots\\ =\dfrac{a}{1-r}\\ =\dfrac{0.9}{1-0.1}\\ =\dfrac{0.9}{0.9}\\ =1$

$0.999999\ldots = 1$ , therefore the statement is $\boxed{\text{True}}$

There's another way to prove this.

Lets divide this 0.999999999999.... by 3. The resultant would be 0.3333333333..... which is $\frac{1}{3}$

$\frac{1}{3}$ Is a third of a one. Therefore if we $multiply$ (0.333333333333..... = $\frac{1}{3}$ ) $by 3$ , the result would be

0.99999999999999...... = 1

Therefore, the statement made above is || True ||

if 0.999999999999999... goes on forever and ever it equals one just as 1.999999999999999999999999999... equals 2!