How do you use a geometric series to prove that #0.999…=1#?

Answer 1
We can write #0.999...=9/10+9/100+9/1000+cdots# #=9/10+9/10(1/10)+9/10(1/10)^2+cdots# #=sum_{n=0}^infty 9/10(1/10)^n#, which is a geometric series with #a=9/10# and #r=1/10#.
So, the sum is #a/{1-r}={9/10}/{1-1/10}=1#
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Answer 2

To prove that (0.999\ldots = 1) using a geometric series, we can represent (0.999\ldots) as an infinite sum of the form:

[0.999\ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \ldots]

This infinite sum forms a geometric series with first term (a = \frac{9}{10}) and common ratio (r = \frac{1}{10}).

The sum of an infinite geometric series is given by the formula:

[S = \frac{a}{1 - r}]

Substituting (a = \frac{9}{10}) and (r = \frac{1}{10}) into the formula, we get:

[S = \frac{\frac{9}{10}}{1 - \frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1]

Therefore, the sum of the infinite geometric series representing (0.999\ldots) is equal to 1, which implies that (0.999\ldots = 1).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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