Infinite Series: Formulas, Examples, and Key Concepts

In Mathematics, a geometric series is the summation of an infinite number of terms that has a constant ratio between each successive term. For example, \[\frac{1}{3}\], \[\frac{1}{9}\], \[\frac{1}{27}\], is a geometric series because each successive term is obtained by multiplying the previous term by \[\frac{1}{3}\]. A geometric series is written in the form of $a + ar +ar^2 + ar^3…$ In the given geometric series, a is the coefficient of each term and r is the common ratio between adjacent terms. 

An infinite geometric series is the summation of an infinite geometric progression or geometric sequence. An infinite geometric series is written in the form of $a_1 + a_1r +a_1r^2 + a_1r^3,$ where $a_1$ is the first term and $r$ is the common ratio between them. An infinite geometric series has a first term and common ratio, but no last term.

Infinite Geometric Series Formula

The formula for the infinite geometric series is given as $S_n = \dfrac{a}{1 - r}$

The first term in the infinite geometric series formula is  $a$’ and the common ratio is $r$. 

Sum of Infinite Geometric Series

An infinite geometric series consists of an infinite number of terms.

The sum of the first n terms, $S_n$, of a geometric series is known as a partial sum.

If $S_n$ approaches to limit as n approaches infinity, the limit is known as the sum to infinity of the series.

The sum of infinite geometric series is given as $Sₙ = \dfrac{a( 1 - r^n)}{1 - r}$

In the above geometric series, the first term is $a$ and the constant ratio is $r$.

When $r > 1, r_n$ approaches infinity as n approaches infinity.

When $r < 1, r_n$ approaches zero as n approaches infinity.

The sum to infinity for a geometric series is undefined when the common ratio $r$ is greater than $1 ( i.e. r > 1)$.

The sum to infinity for a geometric series when $r < 1$ is given as 

$S_{\infty} =  \dfrac{a}{1 - r}$

Example

Find the Sum to Infinity for the Series $96 + 48 + 24$, if it Exists.

Solution:

Here,

$a = 96$, and $r = \dfrac{48}{96} = 0.5$

$S_{\infty} =  \dfrac{a}{1 - r}$

$S_{\infty} = \dfrac{96}{1-0.5}$

$S_{\infty} = \dfrac{96}{0.5}$

$S_{\infty} =192$

Convergent Geometric Series

There is a simple rule to determine if the given geometric series converges or diverges. If $- 1 < r < 1$, then the infinite geometric series will converse whereas if the common ratio r lies outside the interval, then the series will diverge.

Convergent Geometric Series Rule

• The infinite geometric series converges if $- 1 < r < 1$

• The infinite geometric series diverges if $r < - 1\text{ or }r > 1$

The formula to calculate convergent geometric series is given as:

$S_{\infty} = \sum_{i=1}^{n} ar^{i-1} = \dfrac{a( 1 - rⁿ)}{1 - r}$

Let us consider the behaviour of common ratio $r$ for  $- 1 < r < 1$, as the value of n gets larger.

Let $r = \dfrac{1}{3}$

$n = 1, r^n = r^1 = \left(\dfrac{1}{3}\right)^1 = \dfrac{1}{3} \\ n = 1, r^n = r^2 = \left(\dfrac{1}{3}\right)^2 = \dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{9} < \dfrac{1}{3} \\ n = 1, r^n = r^3 = \left(\dfrac{1}{3}\right)^3 = \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{27} < \dfrac{1}{9} $

As $r$ is in the range of $- 1 < r < 1$, we can observe that $r^n$ gets closer to $0$ as $n$ gets larger. Hence $(1 - r^n)$ gets closer to $1$.

Therefore, $S_n = \dfrac{a( 1 - r^n)}{1 - r}$ 

If $- 1 < r < 1$, then $r^n \to 0$ as $n \to \infty$

Therefore ,  $S_{\infty} = \dfrac{a( 1 - 0)}{1 - r} = \dfrac{a}{1 - r}$

The sum of an infinite geometric series formula is given as :

$S_{\infty} = \sum_{i = 1}^{\infty} ar^{i-1} = \dfrac{a}{1 - r} = (- 1 < r < 1)$

Geometric Progression

A geometric progression, also known as the geometric sequence is a sequence of numbers in each term after the first term is calculated by multiplying the previous term by a fixed non-zero number known as the common ratio $r$. 

For example, the sequence $3, 6, 12, 24$ is a geometric progression with a common ratio $2$.

Sum of Infinite GP

The infinite GP sum whose first term is a and common ratio $r$ $(-1 < r < 1,\text{ i.e. }|r|)$ is given as 

$S_{\infty} = \dfrac{a}{1 -r}$ 

Proof

Let us consider the infinite geometric progression with first term $a$ and common ratio $r$, where $(-1 < r < 1,\text{ i.e. }|r|< 1)$. Hence, the sum of infinite terms in GP, in this case, is given as:

$S_{\infty} = \dfrac{a( 1 - r^n)}{1 - r} = \dfrac{a}{1 - r}  - \dfrac{ar^n}{1 -r} …………(1)$

As $-1 < r < 1$, therefore $r^n$ decreases as the value of $n$ increases and $r^n$ approaches to $0$ as $n$ approaches to infinity i.e. $r^n \to 0$ as $n \to \infty$.

Hence, 

$\dfrac{ar^n}{1 -r} \to 0$ as $n \to \infty$.

Hence from equation $(1)$, the formula for the infinite GP sum is given as :

$S = \lim_{x \to 0} S_n = \lim_{x \to \infty}\left(\dfrac{a}{1 - r}  - \dfrac{ar^2}{1 - r}\right) = \dfrac{a}{1 -r}$ if $|r| < 1$

Note: If $r > 1$, the sum of an infinite GP tends to infinity.

The sum of a geometric series formula can also be used to convert a decimal to a fraction.

$0.6666… = \dfrac{a}{1 - r}$

$=\dfrac{\dfrac{6}{10}}{1-\dfrac{1}{10}}\\ = \dfrac{\dfrac{6}{10}}{\dfrac{9}{10}} \\ = \dfrac{6}{10} \times \dfrac{10}{9} \\ = \dfrac{6}{9} \\ = \dfrac{2}{3}$

Solved Example

1. Find the Sum to Infinity of GP $ -\dfrac{5}{4}, \dfrac{5}{16}, - \dfrac{5}{64}, - \dfrac{5}{256}$

Solution:

The given geometric progression is $ -\dfrac{5}{4}, \dfrac{5}{16}, - \dfrac{5}{64}, - \dfrac{5}{256} …$ Here, the first term is $- \dfrac{5}{4}$ and the common ratio is $ -\dfrac{1}{4}$. Also $|r| < 1$.

Hence, the sum to infinity of GP is calculated as:

$S_\infty = \dfrac{a}{1 -r}  = \dfrac{\dfrac{5}{4}}{1 - \left(-\dfrac{1}{4}\right)} = -1$

2. Given the Geometric Series $18 + 6 + 2$. Find the Sum to Infinity if it Exists.

Solution:

Step 1:

Find the value of the common ratio $(r)$

We need to find the value of $r$ to determine whether the given series converges or diverges.

$\dfrac{a_1}{a_2} = \dfrac{6}{18} = \dfrac{1}{3} \\ \dfrac{a_3}{a_2} = \dfrac{2}{6} = \dfrac{1}{3}$

As $-1 < r < -1$, we can say the given series is a convergent geometric series.

Step 2:

Find the sum to infinity of geometric series

$a = 18, r = \dfrac{1}{3}$

Using the formula:

$S_\infty = \dfrac{a}{1 -r}$ 

$S_\infty = \dfrac{18}{1 - \dfrac{1}{3}} \\ S_\infty = \dfrac{18}{\dfrac{2}{3}} \\ S_\infty = 18 \times \dfrac{3}{2} = 27$

As n approaches infinity, the sum of the series approaches $27$. The value of $n$ will never be greater than $0$, regardless of the number of terms added together.