AP Calculus BC

Does the Sum of 1/10^n Converge? Yes, to 1/9

Q: Does 0.999... really equal 1?

Yes. It is the geometric series \sum_{n=1}^{\infty} 9/10^{n}, whose sum is 9 \cdot \frac{1}{9} = 1 exactly. There is no gap; the two notations name the same number.

The series converges to exactly 1/9. It is geometric with common ratio 1/10, and it is the series the repeating decimal 0.111 and so on stands for, which is why every repeating decimal is a rational number.

\sum_{n=1}^{\infty}\frac{1}{10^{n}}

Converges

\text{sum} = \frac{1}{9}

Settled by the geometric series test.

Why this one is worth knowing

Written out, the partial sums are $0.1$ , $0.11$ , $0.111$ , and so on. The series IS the repeating decimal $0.\overline{1}$ , and its sum being $1/9$ is the proof that the decimal equals a fraction.

The same argument turns any repeating decimal into a fraction: $0.\overline{3} = \sum 3/10^{n} = 3 \cdot \frac{1}{9} = \frac{1}{3}$ , and $0.\overline{9} = 9 \cdot \frac{1}{9} = 1$ , which is a genuine equality rather than an approximation.

The computation

\frac{1/10}{1 - 1/10} = \frac{1/10}{9/10} = \frac{1}{9}

The ratio is $1/10$ , comfortably inside the interval where a geometric series converges, so the formula applies and the answer is exact.

Not sure which test a series wants?

The Convergence Test Chooser walks the decision in order: nth term first, then geometric and p-series pattern matching, then alternating structure, then the ratio test, and finally the comparison family.

Frequently asked questions

Does 0.999... really equal 1?

Yes. It is the geometric series $\sum_{n=1}^{\infty} 9/10^{n}$ , whose sum is $9 \cdot \frac{1}{9} = 1$ exactly. There is no gap; the two notations name the same number.

How do I turn a repeating decimal into a fraction?

Write it as a geometric series and sum it. A block of k repeating digits gives ratio $1/10^{k}$ , so the digits divided by $10^{k} - 1$ is the fraction.