Lesson Explainer: Sum of an Infinite Geometric Sequence | Nagwa Lesson Explainer: Sum of an Infinite Geometric Sequence | Nagwa

Lesson Explainer: Sum of an Infinite Geometric Sequence Mathematics

In this explainer, we will learn how to calculate the sum of an infinite geometric sequence.

A geometric sequence is a sequence that has a common ratio between consecutive terms. We are able to calculate the value of the common ratio by dividing any term by the term that precedes it.

For instance, the following sequence is geometric: 1,3,9,27,81,….

This sequence has a common ratio of 3 since each term can be calculated by multiplying the previous term by 3.

If we define the first term in the general geometric sequence as 𝑇, and the common ratio as π‘Ÿ, we create the following sequence: 𝑇,π‘‡π‘Ÿ,π‘‡π‘Ÿ,π‘‡π‘Ÿ,π‘‡π‘Ÿβ€¦.οŠͺ

We now notice that the exponent of π‘Ÿ in each term is one less than the term number, giving us a general term π‘‡π‘ŸοŠοŠ±οŠ§.

Look at what happens when we divide a term by the term that precedes it: π‘‡π‘Ÿπ‘‡=π‘‡π‘Ÿπ‘‡π‘Ÿ=π‘Ÿ.

No matter which pair of terms we choose, their quotient is always π‘Ÿ, the common ratio.

Let’s generalize this.

Definition:

A geometric sequence is a sequence that has a common ratio between consecutive terms. The general term, π‘‡οŠ, of a geometric sequence with first term 𝑇 and common ratio π‘Ÿ is given by, 𝑇=π‘‡π‘Ÿ.

A geometric series is the sum of a given number of terms of a geometric sequence. A series can be finite or infinite.

Definition:

The common ratio, π‘Ÿ, of a geometric sequence whose 𝑛th term is π‘‡οŠ is given by, π‘Ÿ=𝑇𝑇.

Alternatively, it can be also given by π‘Ÿ=𝑇𝑇.

Now, let’s go back to our earlier example of a geometric sequence: 1,3,9,27,81,….

We notice that as the term number, 𝑛, increases, the value of the term itself, π‘‡οŠ, grows exponentially larger. We might infer, then, that if we were to calculate the sum of a large number of terms, our result would be particularly large. In fact, as 𝑛 approaches infinity for this sequence, the sum of the terms, π‘†οŠ, will also approach infinity.

This is not always the case though. In fact, somewhat counterintuitively, some infinite geometric sequences do have a finite sum. We might see these sorts of sequences when considering fractal geometry, such as calculating the area of a Koch snowflake, or when converting recurring decimals to their equivalent fractional form.

When an infinite geometric sequence has a finite sum, we say that the series (this is just the sum of all the terms) is convergent. In order for a geometric series to be convergent, we need the successive terms to get exponentially smaller until they approach zero. For this to happen, the common ratio must be in the interval ]βˆ’1,1[.

For instance, the following sequence has a common ratio of 12 and is convergent; as 𝑛 approaches infinity, π‘‡οŠ approaches zero, meaning we can find the sum of the infinite sequence: 8,4,2,1,12,….

Definition:

An infinite geometric series is said to be convergent if the absolute value of the common ratio, π‘Ÿ, is less than 1: |π‘Ÿ|<1.

To find a formula for the sum of the terms in an infinite geometric sequence, let’s first consider the finite geometric series with first term 𝑇 and common ratio π‘Ÿ with 𝑛 terms: 𝑆=𝑇+π‘‡π‘Ÿ+π‘‡π‘Ÿ+π‘‡π‘Ÿ+β‹―+π‘‡π‘Ÿ.

Multiplying this equation by π‘Ÿ gives, π‘Ÿπ‘†=π‘‡π‘Ÿ+π‘‡π‘Ÿ+π‘‡π‘Ÿ+π‘‡π‘Ÿ+β‹―+π‘‡π‘Ÿ.οŠͺ

We can now subtract the second equation from the first and factorize fully. Notice that when we subtract the terms on the right-hand side, most of the terms become zero: π‘†βˆ’π‘Ÿπ‘†=π‘‡βˆ’π‘‡π‘Ÿπ‘†(1βˆ’π‘Ÿ)=𝑇(1βˆ’π‘Ÿ).

Dividing both sides of this equation by 1βˆ’π‘Ÿ, we derive the formula for the sum of the first 𝑛 terms of a geometric series with first term 𝑇 and common ratio π‘Ÿ: 𝑆=𝑇(1βˆ’π‘Ÿ)1βˆ’π‘Ÿ.

We stated earlier that for a convergent geometric series, βˆ’1<π‘Ÿ<1.

This means that as 𝑛 approaches infinity, π‘ŸοŠ must approach zero.

In other words, if |π‘Ÿ|<1, then limοŠβ†’βˆžοŠπ‘Ÿ=0.

We can consider what happens with our convergent geometric series as 𝑛 approaches infinity. For |π‘Ÿ|<1, limοŠβ†’βˆžοŠπ‘‡(1βˆ’π‘Ÿ)1βˆ’π‘Ÿ=𝑇(1βˆ’0)1βˆ’π‘Ÿ=𝑇1βˆ’π‘Ÿ.

This is sometimes called the sum to infinity of a geometric series.

Definition: The Sum of an Infinite Geometric Sequence

If the common ratio, π‘Ÿ, satisfies |π‘Ÿ|<1, then the sum of an infinite geometric series with first term 𝑇 is 𝑆=𝑇1βˆ’π‘Ÿ.∞

Let’s now look at a question that requires us to apply our knowledge of common ratios in geometric sequences and conditions for convergence of these series, and calculate the value of a convergent infinite geometric series.

Example 1: Finding the Sum of an Infinite Geometric Series

Find the sum of the geometric series 132+134+138+β‹―.

Answer

We know that if the common ratio, π‘Ÿ, satisfies |π‘Ÿ|<1, then the sum of an infinite geometric sequence with first term 𝑇 is 𝑆=𝑇1βˆ’π‘Ÿ.∞

We can see that the first term is 132, so we will need to calculate the common ratio, π‘Ÿ. We find this by dividing a term by the term that precedes it, so we will use the first two terms: π‘Ÿ=134Γ·132=12.

We can see that the absolute value of the common ratio is less than 1 and so we can find the sum of this series by letting 𝑇=132 and π‘Ÿ=12: 𝑆=1βˆ’=132Γ·12=13.∞

The sum of the series is 13.

In our next example, we will see how to apply this technique when dealing with radical ratios.

Example 2: Determining the Common Ratio of an Infinite Geometric Sequence and Finding Its Sum If it Exists

Consider the series 160+160√2+80+80√2+40+40√2+β‹―.

The series is geometric. What is its common ratio?

Is this series convergent? If yes, what is its sum?

Answer

Part 1

The common ratio in a geometric sequence, π‘Ÿ, is found by dividing a term in the series by the term that precedes it. Let’s choose the first two terms: 160√2Γ·160=1√2.

The common ratio is 1√2.

Note that we would get the same result if we divided the third term by the second, or indeed any term by the term that precedes it!

Part 2

A geometric series is convergent if |π‘Ÿ|<1, or βˆ’1<π‘Ÿ<1.

In this case, βˆ’1<1√2<1, meaning that this series is convergent. We can therefore find the sum of the series with first term 𝑇 and common ratio π‘Ÿ by applying the formula 𝑆=𝑇1βˆ’π‘Ÿβˆž with 𝑇=160 and π‘Ÿ=1√2: 𝑆=1601βˆ’.∞√

To simplify 1βˆ’1√2, we create a common denominator of √2: 1βˆ’1√2=√2√2βˆ’1√2=√2βˆ’1√2.

The sum of the series is now 𝑆=160=160Γ—βˆš2√2βˆ’1=160√2√2βˆ’1.∞√√

To finish, we must remember to rationalize the denominator by multiplying by the conjugate of √2βˆ’1. The conjugate is found by changing the sign between the two terms: 𝑆=160√2√2βˆ’1Γ—βˆš2+1√2+1=160√2ο€»βˆš2+1ο‡ο€»βˆš2βˆ’1ο‡ο€»βˆš2+1=320+160√21.∞

Factorizing this expression, we find 𝑆=160ο€»2+√2.∞

Yes, the series is convergent, with an infinite sum of 160ο€»2+√2.

In our previous two examples, we established the existence of a sum and we calculated that sum based on the first few terms of the series. We can also use a formula for the 𝑛th term of a geometric sequence to achieve the same result.

Example 3: Finding the Sum of an Infinite Number of Terms of a Geometric Sequence given Its General Term

Find the sum of the terms of the infinite geometric sequence starting at π‘‡οŠ§ with 𝑛th term 𝑇=3Γ—14.

Answer

The general term of a geometric series with first term 𝑇 and common ratio π‘Ÿ is 𝑇=π‘‡π‘Ÿ.

Comparing this to our sequence, we see that they do not quite match. Instead, we can use the 𝑛th term formula we were given to generate the first two terms.

When 𝑛=1, 𝑇=3Γ—14=3Γ—14=3.

When 𝑛=2, 𝑇=3Γ—14=3Γ—14=314.

The first term is therefore 3, and the common ratio is 314Γ·3=114.

Since the common ratio is in the interval (βˆ’1,1), the series is convergent and so we can find its sum by using the formula 𝑆=𝑇1βˆ’π‘Ÿβˆž with 𝑇=3 and π‘Ÿ=114: 𝑆=31βˆ’=3Γ·1314=4213.∞οŠͺ

As previously mentioned, the application of this process goes beyond just a given series. We can, in fact, represent a recurring decimal as a fraction by thinking about the decimal as a geometric series.

Example 4: Recurring Decimals

By finding the sum of an infinite geometric sequence, express 0.Μ‡37Μ‡5 as a common fraction.

Answer

The recurring decimal 0.Μ‡37Μ‡5=0.375375375375….

This means we can split it into 0.375+0.000375+0.000000375+β‹― and then write each term as a fraction: 0.Μ‡37Μ‡5=3751000+3751000000+3751000000000+β‹―.

This is a geometric series with first term 3751000 and common ratio 11000. Since the common ratio is in the interval ]βˆ’1,1[, we can say that this series is convergent and thus find its sum.

Using the formula 𝑆=𝑇1βˆ’π‘Ÿβˆž with 𝑇=3751000 and π‘Ÿ=11000 gives 𝑆=1βˆ’=3751000Γ·9991000=375999.∞

Simplifying fully, we see that the recurring decimal 0.Μ‡37Μ‡5 is equivalent to 125333.

Let’s now consider how this process would differ for a recurring decimal whose digits do not all recur.

Example 5: Recurring Decimals

By finding the sum of an infinite geometric sequence, express 0.4Μ‡3 as a common fraction.

Answer

The recurring decimal 0.4Μ‡3=0.4333333….

This means we can split it into 0.4+0.0Μ‡3=0.4+0.03+0.003+0.0003+β‹―.

By considering the sum 0.03+0.003+0.0003+β‹―, we can see that we have a geometric series with a first term 𝑇=0.03. The common ratio is 0.0030.03=110.

Since the absolute value of this common ratio is less than 1, this series is convergent and so we can find its sum.

Using the formula 𝑆=𝑇1βˆ’π‘Ÿβˆž with 𝑇=0.03 and π‘Ÿ=110 gives 𝑆=0.031βˆ’=0.03Γ·910=390.∞

Simplifying fully, we see that that the recurring decimal 0.0Μ‡3 is equivalent to 130.

This means that 0.4Μ‡3=0.4+130=1330.

As a common fraction, 0.4Μ‡3 is 1330.

In our next example, we will look at how to find the infinite sum of a geometric series given the value of two of its terms. This will involve applying the formula for the general term of a geometric sequence and then working backward to calculate the value of the common ratio.

Example 6: Finding the Sum of an Infinite Geometric Sequence given the Values of Two Terms

Find the sum of an infinite geometric sequence given the first term is 171 and the fourth term is 17164.

Answer

A geometric series is convergent if |π‘Ÿ|<1, or βˆ’1<π‘Ÿ<1, where π‘Ÿ is the common ratio.

In this case, the sum of an infinite geometric sequence with first term 𝑇 is 𝑆=𝑇1βˆ’π‘Ÿ.∞

Notice that we have been given the value of the first and fourth terms, so we will need to use this information to calculate the common ratio.

We will use the formula for the 𝑛th term of a geometric sequence with 𝑇=171 and 𝑇=17164οŠͺ: 𝑇=π‘‡π‘Ÿ17164=171π‘Ÿ17164=171π‘Ÿ.οŠͺ

To solve for π‘Ÿ, we will divide through by 171 and find the cube root of both sides of the equation: 164=π‘Ÿπ‘Ÿ=14.

Since the absolute value of this common ratio is less than 1, this series is convergent and so we can find its sum.

Using the formula 𝑆=𝑇1βˆ’π‘Ÿβˆž with 𝑇=171 and π‘Ÿ=14 gives 𝑆=1711βˆ’=171Γ·34=228.∞οŠͺ

The sum of the infinite geometric sequence is 228.

In our final example, we look at how we can apply the formula for the infinite sum of a geometric series to calculate the first term.

Example 7: Finding the First Term of an Infinite Geometric Sequence given Its Common Ratio and the Sum of the Terms

Find the first term of the infinite geometric sequence where the common ratio is 14 and the sum is 9867.

Answer

We know that if the common ratio, π‘Ÿ, satisfies |π‘Ÿ|<1, then the sum of an infinite geometric sequence with first term 𝑇 is 𝑆=𝑇1βˆ’π‘Ÿ.∞

Letting π‘Ÿ=14, we see that the absolute value of π‘Ÿ does indeed satisfy the requirement for a series to be convergent.

Letting 𝑆=9867∞, our formula for the sum to infinity becomes 9867=𝑇1βˆ’9867=𝑇.οŠͺοŠͺ

To solve for 𝑇, the first term, we multiply both sides of this equation by 34: 𝑇=9867Γ—34=6927Γ—34=5197.

The first term of the infinite geometric sequence is 5197.

Key Points

  • An infinite geometric series is said to be convergent if the absolute value of the common ratio, π‘Ÿ, is less than 1: |π‘Ÿ|<1.
  • For a convergent geometric series with first term 𝑇, the infinite sum is given by 𝑆=𝑇1βˆ’π‘Ÿ.∞
  • By expressing a recurring decimal as a geometric sequence, we can find its sum and write it as a common fraction.

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