MethodMath
Nova AI
Apr 28, 2026

Does the infinite series 9 divided by 10 to the power of n starting from 1 converge to 1 proof

For the 0.999...=10.999... = 1 question, I want to see a rigorous proof using infinite series.

The decimal 0.999...0.999... can be written as:

0.9=n=1910n=910+9100+91000+0.\overline{9} = \sum_{n=1}^{\infty} \frac{9}{10^n} = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots

This is a geometric series with a=910a = \frac{9}{10} and r=110r = \frac{1}{10}.

The sum is a1r=9/1011/10=9/109/10=1\frac{a}{1-r} = \frac{9/10}{1-1/10} = \frac{9/10}{9/10} = 1.

But someone told me this only works for "convergent" series. What does that mean? And isnt there a number between 0.999... and 1? By the density of real numbers, shouldnt there be?

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1 Answer

Ahmed Al-Rashid
Ahmed Al-RashidMay 22, 2026
For limits, I would first classify the expression before applying any rule. If direct substitution gives a number, you are done. If it gives $0/0$ or $\infty/\infty$, then algebra, series expansion, or L'Hopital's rule may be useful. If it gives something like $0\cdot\infty$ or $\infty-\infty$, rewrite it into a quotient first. For example, near $0$ the expansion $$\sin x=x-\frac{x^3}{6}+O(x^5)$$ often gives the cleanest answer. It also explains why repeated L'Hopital works in some cases but can hide the structure of the expression.
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Dr. Aisha Mohammed
Dr. Aisha MohammedMay 23, 2026

This helped. The check at the end is the part I usually forget.

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