Geometric Series

Definition

1.A geometric series is based on a geometric sequence, where each term is obtained by multiplying the previous term by a constant ratio. It is defined as the following sum:

$$ \sum_{n=0}^{\infty} r^n $$

2.The value $r$ is the constant factor used to multiply each term to produce the next. If $r = 0$, the series converges to $1$, because all subsequent terms are zero: $$ \sum_{n=0}^{\infty} 0^n = 1 + 0 + 0 + \dots = 1 $$

3.If $r = 1$, the series diverges, because the sequence of partial sums grows without bound:

$$ s_n = 1 + 1 + \dots + 1 = n + 1 \quad \rightarrow \quad \lim_{n \to \infty} s_n = \infty $$

4.The geometric series serves as a reference model for analyzing more complex series, because its behavior is well understood and defined. For series with positive terms, it is used in comparison tests to determine convergence or divergence. If a given series can be bounded above or below by a geometric series, we can often deduce the nature of the series by analogy.

5.The $n$-th partial sum of a geometric series with $r \ne 0$ and $r \ne 1$ is given by:

$$ S_n = \sum_{k=0}^{n} r^k = \frac{1 - r^{n+1}}{1 - r} $$

6.This result is obtained starting from:

$$ s_n = 1 + r + r^2 + r^3 + \dots + r^n $$

7.Multiplying both sides by $r$, we get:

$$ \begin{align} r \cdot s_n &= r(1 + r + r^2 + \dots + r^n) \\[6pt] &= r + r^2 + r^3 + \dots + r^n + r^{n+1} \\[6pt] &= 1+ r + r^2 + r^3 + \dots + r^n + r^{n+1} -1 \\[6pt] \end{align} $$

8.In the last term of the equation, we added and subtracted $+1$. This way, the expression from $1 + \dots + r^n$ becomes $s_n$. By substituting the values, we obtain:

$$ \begin{align} r \cdot s_n &= s_n + r^{n+1} - 1 \\[6pt] s_n (1 - r) &= 1 - r^{n+1} \end{align} $$

9.and finally:

$$ s_n = \frac{1 - r^{n+1}}{1 - r} \quad \text{for } r \ne 1 $$

10.

The condition $r \ne 1$ is essential; otherwise, the denominator becomes zero and the formula is undefined.

11.When $|r| < 1$ the geometric series converges, because the terms decrease in absolute value. The infinite sum is computed using the formula:

$$ \sum_{n=0}^{\infty} r^n = \frac{1}{1 - r} $$

12.This result holds because $\lim_{n \to \infty} r^{n+1} = 0$ when $|r| < 1$, which simplifies the formula for the partial sum:

$$ s_n = \frac{1 - r^{n+1}}{1 - r} \quad \rightarrow \quad \lim_{n \to \infty} s_n = \frac{1}{1 - r} $$

13.A geometric series may have a first term different from $1$. Multiplying every term by a constant $a$ gives the general form:

$$ \sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + \dots $$

14.The constant $a$ factors out of the sum, so the partial sum follows from the case with first term $1$:

$$ S_n = a\frac{1 - r^{n+1}}{1 - r} $$

15.When $|r| < 1$ the series converges, and its sum is:

$$ \sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r} $$

16.A nonzero constant $a$ scales every term by the same factor, so it changes the value of the limit but not whether the series converges. In general, when the index starts at a value $n > 0$, for example $\alpha$, the formula for the sum of the geometric series becomes:

$$ \sum_{n=\alpha}^{\infty} r^n = \frac{r^{\alpha}}{1 - r} $$

17.When $r > 1$ the geometric series diverges, because the terms $r^n$ grow without bound as $n \to \infty$, and so does the sequence of partial sums:

$$ \lim_{n \to \infty} s_n = \lim_{n \to \infty} \sum_{k=0}^{n} r^k = \infty $$

18.Since the terms do not tend to zero and their sum grows without bound, the series does not have a finite limit and therefore diverges.

19.When $r \leq -1$ the limit of $s_n$ does not exist, and the series is indeterminate. The terms $r^n$ alternate in sign and grow in magnitude, so the partial sums $s_n$ oscillate without approaching any finite value, and the series neither converges nor diverges to infinity. For example, if $r = -2$, the terms become:

$$ 1 - 2 + 4 - 8 + 16 - \dots $$

20.and the partial sums fluctuate without stabilizing.

21.A different behavior appears at the boundary $r = -1$, where the terms keep a constant magnitude instead of growing:

$$ 1 - 1 + 1 - 1 + \dots $$

22.The partial sums alternate between $1$ and $0$, so they stay bounded yet still fail to approach a single value. The series is indeterminate, like every case with $r \leq -1$.

23.Therefore, we can conclude that the series:

Relation to power series

25.Replacing the common ratio with a variable $x$ turns the geometric series into a power series:

$$ \sum_{n=0}^{\infty} x^n = \frac{1}{1 - x} \quad \text{for } |x| < 1 $$

26.This is the simplest power series, the one whose coefficients are all equal to $1$, and the condition $|x| < 1$ gives it a radius of convergence equal to $1$. The same identity supports the ratio test and the root test, which establish the convergence of a general series by comparing it to a geometric one.

Example

27.We study the following geometric series and compute its sum.

$$\sum_{n=0}^{\infty} \frac{3^n + 4^n}{5^n}$$

28.By the linearity of summation, we can rewrite the series as the sum of two geometric series:

$$\sum_{n=0}^{\infty} \left( \frac{3}{5} \right)^n + \sum_{n=0}^{\infty} \left( \frac{4}{5} \right)^n$$

29.The common ratio of both series is $< 1$, so by the properties of geometric series both converge. The sum of a geometric series is given by:

$$ \sum_{n=0}^{\infty} r^n = \frac{1}{1 - r} $$

30.We have:

$$\sum_{n=0}^{\infty} \left( \frac{3}{5} \right)^n = \frac{1}{1 - \frac{3}{5}} = \frac{1}{\frac{2}{5}} = \frac{5}{2}$$

$$\sum_{n=0}^{\infty} \left( \frac{4}{5} \right)^n = \frac{1}{1 - \frac{4}{5}} = \frac{1}{\frac{1}{5}} = 5$$

31.By adding the two values, we obtain:

$$\frac{5}{2} + 5 = \frac{15}{2}$$

32.Therefore, the series converges and its sum is: $$ \frac{15}{2} $$

Repeating decimals

33.A decimal with an infinitely repeating pattern can be written as a geometric series, which turns it into a ratio of two integers. Consider the number $0.\overline{12} = 0.121212\dots$, where the block $12$ repeats without end. It can be expressed as:

$$ 0.\overline{12} = \sum_{n=1}^{\infty} \frac{12}{100^n} $$

34.This is a geometric series with first term $a=\frac{12}{100}$ and common ratio $r=\frac{1}{100}$. Since $|r| < 1$, the series converges, and applying the formula for the sum gives:

$$ 0.\overline{12} = \frac{\frac{12}{100}}{1 - \frac{1}{100}} = \frac{\frac{12}{100}}{\frac{99}{100}} = \frac{12}{99} = \frac{4}{33} $$

35.The same procedure rewrites any repeating decimal as a fraction, which shows that every such number is rational.