Infinite Geometric Series and Convergence

Definition The infinite geometric series with first term $a_1 \ne 0$ and common ratio $r \ne 0$ is

Its $n$th partial sum is the sum of its first $n$ terms,

The value of the infinite series, if it has one, is defined through the sequence $S_1, S_2, S_3, \dots$ of partial sums.

Definition A series $a_1 + a_2 + a_3 + \cdots$ converges to the number $S$, called its sum, if its partial sums $S_n = a_1 + a_2 + \cdots + a_n$ approach $S$ as $n$ grows. Precisely, for every positive number $\varepsilon$ there is an integer $N$ such that

In that case one writes $a_1 + a_2 + a_3 + \cdots = S$. For a geometric series the terms are $a_n = a_1 r^{n-1}$.

Definition A series that does not converge diverges. Its partial sums may tend to $\infty$ or to $-\infty$, or they may keep moving around without settling near any single number. Since each term after the first is the difference $a_n = S_n - S_{n-1}$ of two partial sums, a series whose partial sums approach $S$ has terms that approach $S - S = 0$. So a series whose terms do not approach $0$ diverges. The converse is false. Terms that approach $0$ do not on their own make a series converge.

Definition For a real number $r$, the powers $r^n$ behave as follows as $n$ grows.

• If $\lvert r \rvert < 1$, then $r^n \to 0$.
• If $r = 1$, then $r^n = 1$ for every $n$.
• If $r = -1$, then $r^n$ alternates between $-1$ and $1$.
• If $\lvert r \rvert > 1$, then $\lvert r^n \rvert$ grows without bound.

For $0 < \lvert r \rvert < 1$ the first fact has a direct reason. Given a tolerance $\varepsilon$ with $0 < \varepsilon < 1$, take logarithms in $\lvert r \rvert^n < \varepsilon$ and divide by the negative number $\ln \lvert r \rvert$, which reverses the inequality. So $\lvert r^n \rvert < \varepsilon$ exactly when

Definition Let $a_1 \ne 0$. The geometric series $a_1 + a_1 r + a_1 r^2 + \cdots$ converges if and only if $\lvert r \rvert < 1$, and in that case

The reason is the partial sum formula. For $\lvert r \rvert < 1$ it can be written

and the last term approaches $0$ because $r^n \to 0$. For $\lvert r \rvert \ge 1$ every term has size $\lvert a_1 \rvert \lvert r \rvert^{n-1} \ge \lvert a_1 \rvert > 0$, so the terms do not approach $0$ and the series diverges. The calculus lessons start the index at $0$ and write the same series as $\sum_{k=0}^{\infty} a_1 r^k$, with the same sum.

Definition Let $a_1 \ne 0$. For $r = 1$ the series is $a_1 + a_1 + a_1 + \cdots$ with partial sums $S_n = n\,a_1$, which tend to $\infty$ if $a_1 > 0$ and to $-\infty$ if $a_1 < 0$. For $r = -1$ the series is $a_1 - a_1 + a_1 - \cdots$ and its partial sums alternate,

Both series diverge, and the formula $\frac{a_1}{1 - r}$ applies to neither of them.

Definition When a block of $k$ digits, not all zero, repeats forever in a decimal, the repeating part is an infinite geometric series with ratio $10^{-k}$. If the block starts right after the decimal point and its digits form the whole number $d$, then

Since $0 < 10^{-k} < 1$, the series converges. Leading zeros count as digits of the block, so the block $07$ has $k = 2$ and $d = 7$, and $0.070707\ldots = \frac{7}{99}$. A decimal whose repeating block starts later is a terminating decimal plus such a value divided by a power of $10$, so every repeating decimal is a rational number.