# How can I tell whether a geometric series converges?

Jul 2, 2015

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 $1$.