The sum to infinity of a geometric sequence can be calculated when the common ratio is a number less than 1 and greater than -1. For this, we simply need the value of the first term and the value of the common ratio. We then use these values in a standard formula.

Here, we will explore the formula that we can use to find the sum to infinity of a geometric sequence. Then, we will apply this formula to solve some exercises.

## Formula for the sum to infinity of geometric sequence

The sum to infinity of a geometric sequence can be calculated when we have $latex -1<r<1$. For example, consider the following sequence:

$$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+…\left(\frac{1}{2}\right)^n$$

We can see that as the number of terms increases, the value of each term gets smaller and smaller.

That is, as $latex n \rightarrow \infty$, the term $latex \left(\frac{1}{2}\right)^n\rightarrow 0$, therefore the sum to infinity of the sequence has a specific value.

To find the sum to infinity of a geometric sequence, we use the following formula:

$$S_{\infty}= \frac{a}{1-r}$$

where $latex -1<r<1$. If the common ratio doesn’t meet this condition, the infinite sum does not exist.

## Proof of the formula for the sum to infinity of geometric sequences

Recall that we can find the sum of the first $latex n$ terms of any geometric sequence using the following formula:

$$S_{n}=\frac{a(1-r^n)}{1-r}$$

If the common ratio satisfies the condition $latex -1<r<1$, as $latex n\rightarrow \infty$, we have $latex r^n\rightarrow 0$.

Basically, this means that the $latex r^n$ term of the progression gets smaller and smaller until its value gets closer and closer to 0.

Therefore, as $latex n\rightarrow \infty$, we have:

$$S_{n}\rightarrow \frac{a(1-0)}{1-r}=\frac{a}{1-r}$$

Therefore, we prove that the sum to infinity of a geometric sequence where $latex -1<r<1$ is given by:

$$S_{\infty}=\frac{a}{1-r}$$

We note that the proof begins by assuming that $latex -1<r<1$. If this is not the case, the sum to infinity does not exist.

## Examples of the sum to infinity of geometric sequences

**EXAMPLE **1

**EXAMPLE**

Find the sum to infinity of the following geometric sequence:

$$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+…$$

##### Solution

We start by writing the information we know:

- First term: $latex a=1$
- Common ratio: $latex r=\frac{1}{2}$

Now, we use the formula for the sum to infinity of a geometric sequence:

$$S_{\infty}=\frac{a}{1-r}$$

$$=\frac{1}{1-\frac{1}{2}}$$

$$=\frac{1}{\frac{1}{2}}$$

$latex S_{\infty}=2$

**EXAMPLE **2

**EXAMPLE**Find the sum to infinity of a geometric sequence in which the first term is equal to -3 and the common ratio is equal to $latex -\frac{1}{2}$.

##### Solution

In this case, we have the following values:

- First term: $latex a=-3$
- Common ratio: $latex r=-\frac{1}{2}$

By applying the formula for the sum to infinity of a geometric sequence, we have:

$$S_{\infty}=\frac{a}{1-r}$$

$$=\frac{-3}{1+\frac{1}{2}}$$

$$=\frac{-3}{\frac{3}{2}}$$

$latex S_{\infty}=-2$

**EXAMPLE **3

**EXAMPLE**Find the sum to infinity of the following geometric sequence:

$$2+\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+…$$

##### Solution

In this case, we have the following

- First term: $latex a=2$
- Common ratio: $latex r=\frac{1}{4}$

When we use the formula for the sum to infinity, we have:

$$S_{\infty}=\frac{a}{1-r}$$

$$=\frac{2}{1-\frac{1}{4}}$$

$$=\frac{2}{\frac{3}{4}}$$

$$S_{\infty}=\frac{8}{3}$$

**EXAMPLE **4

**EXAMPLE**Find the result of the following sum:

$$\sum_{r=1}^{\infty}\left(\frac{1}{5}\right)^r$$

##### Solution

Here, we have the sum of the geometric sequence written in sigma notation. We find the value of the first term by using $latex r=1$.

The common ratio is the number inside the parentheses, so we have:

- First term: $latex a=\frac{1}{5}$
- Common ratio: $latex r=\frac{1}{5}$

Using the formula for the sum to infinity with these values, we have:

$$S_{\infty}=\frac{a}{1-r}$$

$$=\frac{\frac{1}{5}}{1-\frac{1}{5}}$$

$$=\frac{\frac{1}{5}}{\frac{4}{5}}$$

$$=\frac{5}{20}$$

$$S_{\infty}=\frac{1}{4}$$

**EXAMPLE **5

**EXAMPLE**Write the recurring decimal 0.3232… as the sum of a geometric sequence, and then write the decimal as a rational number.

##### Solution

To solve this exercise, we begin by writing the decimal as the sum of a geometric sequence:

$$0.323232…= \frac{32}{100}+\frac{32}{10000}+\frac{32}{1000000}+…$$

Now, we identify the following values:

- $latex a=\frac{32}{100}$
- $latex r=\frac{1}{100}$

Since $latex -1<r<1$, we can calculate the value of the sum to infinity:

$$S_{\infty}=\frac{a}{1-r}$$

$$=\frac{\frac{32}{100}}{1-\frac{1}{100}}$$

$$=\frac{\frac{32}{100}}{\frac{99}{100}}$$

$$S_{\infty}=\frac{32}{99}$$

Therefore, the recurring decimal 0.3232…, can be written as $latex \frac{32}{99}$.

**EXAMPLE **6

**EXAMPLE**Express the recurring decimal 0.7272… as a fraction in its simplest form.

##### Solution

We write to the decimal as a geometric sequence:

$$0.727272…= \frac{72}{100}+\frac{72}{10000}+\frac{72}{1000000}+…$$

Now, we can see that the sequence has the following values:

- $latex a=\frac{72}{100}$
- $latex r=\frac{1}{100}$

Since $latex -1<r<1$, we can calculate the value of the sum to infinity:

$$S_{\infty}=\frac{a}{1-r}$$

$$=\frac{\frac{72}{100}}{1-\frac{1}{100}}$$

$$=\frac{\frac{72}{100}}{\frac{99}{100}}$$

$$=\frac{72}{99}$$

$$S_{\infty}=\frac{8}{11}$$

Therefore, the recurring decimal 0.7272…, can be written as $latex \frac{8}{11}$.

## Sum to infinity of geometric sequences – Practice problems

#### Find the sum to infinity of the following geometric sequence $$4+\frac{4}{3}+\frac{4}{9}+…$$

Write the answer in the input box.

## See also

Interested in learning more about geometric sequences? You can ake a look at these pages:

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