Convergence of Geometric Series

Key Questions

A geometric series of geometric sequence ${u}_{n} = {u}_{1} \cdot {r}^{n - 1}$ converges only if the absolute value of the common factor $r$ of the sequence is strictly inferior to $1$; in other words, if $| r | < 1$.

Explanation:

The standard form of a geometric sequence is :

${u}_{n} = {u}_{1} \cdot {r}^{n - 1}$

And a geometric series can be written in several forms :

${\sum}_{n = 1}^{+ \infty} {u}_{n} = {\sum}_{n = 1}^{+ \infty} {u}_{1} \cdot {r}^{n - 1} = {u}_{1} {\sum}_{n = 1}^{+ \infty} {r}^{n - 1}$

$= {u}_{1} \cdot {\lim}_{n \to + \infty} \left({r}^{1 - 1} + {r}^{2 - 1} + {r}^{3 - 1} + \ldots + {r}^{n - 1}\right)$

Let ${r}_{n} = {r}^{1 - 1} + {r}^{2 - 1} + {r}^{3 - 1} + \ldots + {r}^{n - 1}$

Let's calculate ${r}_{n} - r \cdot {r}_{n}$ :

${r}_{n} - r \cdot {r}_{n} = {r}^{1 - 1} - {r}^{2 - 1} + {r}^{2 - 1} - {r}^{3 - 1} + {r}^{3 - 1} + \ldots - {r}^{n - 1} + {r}^{n - 1} - {r}^{n} = {r}^{1 - 1} - {r}^{n}$

${r}_{n} \left(1 - r\right) = {r}^{1 - 1} - {r}^{n} = 1 - {r}^{n}$

${r}_{n} = \frac{1 - {r}^{n}}{1 - r}$

Therefore, the geometric series can be written as :

${u}_{1} {\sum}_{n = 1}^{+ \infty} {r}^{n - 1} = {u}_{1} \cdot {\lim}_{n \to + \infty} \left(\frac{1 - {r}^{n}}{1 - r}\right)$

Thus, the geometric series converges only if the series ${\sum}_{n = 1}^{+ \infty} {r}^{n - 1}$ converges; in other words, if ${\lim}_{n \to + \infty} \left(\frac{1 - {r}^{n}}{1 - r}\right)$ exists.

• If |r| > 1 : ${\lim}_{n \to + \infty} \left(\frac{1 - {r}^{n}}{1 - r}\right) = \infty$

• If |r| < 1 : ${\lim}_{n \to + \infty} \left(\frac{1 - {r}^{n}}{1 - r}\right) = \frac{1}{1 - r}$.

Therefore, the geometric series of geometric sequence ${u}_{n}$ converges only if the absolute value of the common factor $r$ of the sequence is strictly inferior to $1$.

Here are some examples:

$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots$

$1 - 1 + 1 - 1 + 1 - 1 + \ldots$

$1 + 2 + 4 + 8 + 16 + \ldots$

Explanation:

All geometric series are of the form ${\sum}_{i = 0}^{\infty} a {r}^{i}$ where $a$ is the initial term of the series and $r$ the ratio between consecutive terms.

In the three examples above, we have:

$a = 1$, $r = \frac{1}{2}$

${\sum}_{i = 0}^{\infty} a {r}^{i} = 2$

$a = 1$, $r = - 1$

${\sum}_{i = 0}^{\infty} a {r}^{i}$ does not converge - it alternates between $0$ and $1$ as each term is added.

$a = 1$, $r = 2$

${\sum}_{i = 0}^{n} a {r}^{i} \to \infty$ as $n \to \infty$

The geometric series ${\sum}_{i = 0}^{\infty} a {r}^{i}$ only converges in the following cases:

(1) $a = 0$

${\sum}_{i = 0}^{\infty} a {r}^{i} = 0$

(2) $\left\mid r \right\mid < 1$

${\sum}_{i = 0}^{\infty} a {r}^{i} = \frac{a}{1 - r}$

• $a + a x + a {x}^{2} + a {x}^{3} + \cdots$, which converges to $\frac{a}{1 - x}$ when $| x | < 1$. More generally, you could also write $a + a \left(x - c\right) + a {\left(x - c\right)}^{2} + a {\left(x - c\right)}^{3} + \cdots$, which converges to $\frac{a}{1 - \left(x - c\right)} = \frac{a}{\left(1 + c\right) - x}$ when $| x - c | < 1$.