Infinite Geometric Series

Grade 12 · geometry · 71 practice problems · read aloud

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Infinite Geometric Series

📚 What is it and Why is it Useful?

An infinite geometric series is the sum of an endless sequence of terms where each term is found by multiplying the previous term by a constant, called the common ratio (r). It's a powerful concept in calculus and advanced algebra, used to model real-world phenomena like repeating decimals, fractal patterns, and financial calculations involving perpetual payments.

🧮 Step-by-Step Guide

An infinite geometric series converges (adds up to a finite number) only if |r| < 1. If |r| ≥ 1, the sum diverges (is infinite).

  1. Step 1: Identify the first term, a.
  2. Step 2: Find the common ratio, r.
  3. Step 3: Check the convergence condition: Is |r| < 1? If not, the sum does not exist.
  4. Step 4: Apply the formula: S∞ = a / (1 - r)

🔢 Visual Examples

Example 1: Convergent Series

Find the sum of 8 + 4 + 2 + 1 + ...

  • Step 1: a = 8
  • Step 2: r = 4/8 = 1/2
  • Step 3: |1/2| < 1, so it converges.
  • Step 4: S∞ = 8 / (1 - 1/2) = 8 / (1/2) = 16

Example 2: Fractional Terms

Find the sum of 12 + 3 + 0.75 + 0.1875 + ...

  • Step 1: a = 12
  • Step 2: r = 3/12 = 1/4
  • Step 3: |1/4| < 1, so it converges.
  • Step 4: S∞ = 12 / (1 - 1/4) = 12 / (3/4) = 16

⚠️ Common Mistakes

  • Forgetting the Convergence Check: Always verify |r| < 1 before using the formula. Applying it to a divergent series like 2 + 4 + 8 + ... is a critical error.
  • Misidentifying the First Term: The first term (a) is simply the first number in the sequence, not necessarily the coefficient in a formula.
  • Sign Errors: Be extremely careful with negative common ratios. S∞ = a / (1 - r). If r is negative, you are subtracting a negative number in the denominator.

💡 Tips & Tricks

  • Memory Aid: The formula is "First over One Minus Ratio" (F / OMR).
  • Visualize Shrinking: Imagine the terms getting smaller and smaller, "filling up" a finite space. This helps build intuition for why an infinite sum can be finite.
  • Repeating Decimals: To convert a repeating decimal like 0.444... to a fraction, recognize it as the series 0.4 + 0.04 + 0.004 + ... where a=0.4 and r=0.1. S∞ = 0.4/(1-0.1) = 0.4/0.9 = 4/9.

🎯 Practice Suggestions

To master infinite geometric series:

  1. Start by practicing identifying a and r from long sequences.
  2. Create your own convergent series by choosing an a and an r where |r| < 1.
  3. Practice with word problems, especially those involving bouncing balls or financial investments with perpetual growth/decay.
  4. Always write down the convergence check as the first step in your solution.

Practice problems

6 of the 71, worked through step by step — try them before opening the answer.

1 ∑_{n=1}^{∞} (2/3)^n = ?

Hint: For an infinite geometric series, identify the first term and common ratio. The sum exists only when the absolute value of the ratio is less than 1.

Show the answer

Answer: 2

  1. Identify the type of series. This is a geometric series of the form: a + a r + a r^2 + ... , but here the sum starts at n = 1, not n = 0.
  2. Write the first few terms. For n = 1: (2/3)^1 = 2/3 n = 2: (2/3)^2 = 4/9 n = 3: (2/3)^3 = 8/27 and so on. So the series is: 2/3 + 4/9 + 8/27 + ...
  3. Identify the first term and common ratio. First term a = 2/3. Common ratio r = 2/3 (since each term is multiplied by 2/3 to get the next term).
  4. Use the sum formula for an infinite geometric series. The sum S of an infinite geometric series with first term a and common ratio r (where |r| < 1) is: S = a / (1 - r). But careful: this formula works when the sum starts at n = 0. Our sum starts at n = 1.
  5. Adjust for starting index. If the sum were from n = 0 to infinity: sum = a / (1 - r) with a = first term when n = 0. Here, when n = 0, term = (2/3)^0 = 1. So sum from n = 0 to infinity = 1 / (1 - 2/3) = 1 / (1/3) = 3. But our sum is from n = 1 to infinity, so we subtract the n = 0 term (which is 1): Sum from n = 1 to infinity = (sum from n = 0 to infinity) - (term at n = 0) = 3 - 1 = 2.
  6. Alternatively, direct formula for n = 1 to infinity. For sum from n = 1 to infinity of a r^(n-1), the sum is a / (1 - r). But here our general term is a r^n with a = 2/3, r = 2/3. So sum from n = 1 to infinity of a r^n = (a r) / (1 - r) if we factor? Let's check: Sum from n = 1 to infinity of r^n = r + r^2 + r^3 + ... = r (1 + r + r^2 + ...) = r * [1 / (1 - r)] = r / (1 - r). So with r = 2/3: Sum = (2/3) / (1 - 2/3) = (2/3) / (1/3) = (2/3) * (3/1) = 2.
  7. Conclusion. Thus, the sum from n = 1 to infinity of (2/3)^n equals 2. Final answer: 2

We are given the infinite series: sum from n= 1 to infinity of (2/3)^n.

2 ∑_{n=1}^∞ 3(0.6)^{n-1} = ?

Hint: For an infinite geometric series, check if the common ratio's absolute value is less than 1 to determine convergence, then apply the appropriate formula.

Show the answer

Answer: 7.5

  1. Identify the type of series** This is a geometric series of the form: a + a r + a r^2 + a r^3 + ... where the first term a = 3, and the common ratio r = 0.6. Check: For n = 1, term = 3 * (0.6)^(0) = 3 For n = 2, term = 3 * (0.6)^(1) = 1.8 For n = 3, term = 3 * (0.6)^(2) = 1.08 Indeed, each term is multiplied by 0.6 to get the next term. --- **
  2. Condition for convergence** An infinite geometric series converges if |r| < 1. Here r = 0.6, which is less than 1, so the series converges. --- **
  3. Formula for the sum of an infinite geometric series** The sum S of an infinite geometric series with first term a and common ratio r (|r| < 1) is: S = a / (1 - r) --- **
  4. Substitute values** Here a = 3, r = 0.6 S = 3 / (1 - 0.6) S = 3 / 0.4 --- **
  5. Simplify** 3 / 0.4 = 30 / 4 = 15 / 2 = 7.5 --- **
  6. Conclusion** The sum of the infinite series is 7.5. --- **Final answer:** 7.5

We are given the infinite series: sum from n= 1 to infinity of 3 * (0.6)^(n-1) --- **

3 ∑_{n=1}^{∞} 4(0.6)^{n-1} = ?

Hint: For an infinite geometric series, consider the relationship between the first term and the common ratio to determine convergence.

Show the answer

Answer: 10

  1. Recognize the series type** This is a geometric series of the form: a + a r + a r^2 + a r^3 + ... Here, the first term a = 4 * (0.6)^(1-1) = 4 * (0.6)^0 = 4 * 1 = 4. The common ratio r = 0.6. --- **
  2. Check convergence condition** For a geometric series, it converges if |r| < 1. Here, |0.6| = 0.6 < 1, so it converges. --- **
  3. Apply the infinite geometric series sum formula** The sum S of an infinite geometric series with first term a and common ratio r (|r| < 1) is: S = a / (1 - r) Here, a = 4, r = 0.6. So S = 4 / (1 - 0.6) --- **
  4. Calculate step by step** 1 - 0.6 = 0.4 So S = 4 / 0.4 4 / 0.4 = 40 / 4 = 10 --- **
  5. Conclusion** The sum of the infinite series is 10. --- **Final answer:** 10

We are given the infinite series: sum from n= 1 to infinity of 4 * (0.6)^(n-1) --- **

4 ∑(n=1 to ∞) 4(0.6)^(n-1) = ?

Hint: For an infinite geometric series, the sum exists when the absolute value of the common ratio is less than 1. The formula involves the first term and this ratio.

Show the answer

Answer: 10

  1. Identify the type of series** This is a geometric series of the form: a + a r + a r^2 + a r^3 + ... Here, the first term a = 4 * (0.6)^(1 - 1) = 4 * (0.6)^0 = 4 * 1 = 4. The common ratio r = 0.6. --- **
  2. Check the condition for convergence** For an infinite geometric series, it converges if |r| < 1. Here, r = 0.6, which is less than 1, so the series converges. --- **
  3. Apply the sum formula for infinite geometric series** The sum S of an infinite geometric series with first term a and common ratio r (|r| < 1) is: S = a / (1 - r) Here, a = 4, r = 0.6. So S = 4 / (1 - 0.6) --- **
  4. Perform the calculation** 1 - 0.6 = 0.4 So S = 4 / 0.4 4 divided by 0.4 = 40 / 4 = 10. --- **
  5. Conclusion** The sum of the infinite series is 10. --- **Final answer:** 10

We are given the infinite series: Sum from n = 1 to infinity of 4 * (0.6)^(n - 1) --- **

5 ∑_{n=1}^{∞} 5(0.8)^{n-1} = ?

Hint: For an infinite geometric series, consider the relationship between the first term and the common ratio to determine convergence.

Show the answer

Answer: 25

  1. Identify the first term (a) and common ratio (r) from the series ∑_{n=1}^{∞} 5(0.8)^{n-1} The first term a = 5 Common ratio r = 0.8
  2. Check if the series converges Since |r| = |0.8| = 0.8 < 1, the series converges
  3. Apply the infinite geometric series formula Sum = a/(1 - r) Sum = 5/(1 - 0.8)
  4. Calculate the sum Sum = 5/0.2 Sum = 25

The answer is 25.

6 ∑(n=1 to ∞) 5(0.8)^(n-1) = ?

Hint: For an infinite geometric series, the sum exists only if the absolute value of the common ratio is less than 1. The formula involves the first term and this ratio.

Show the answer

Answer: 25

  1. Identify the first term (a) and common ratio (r) The series is 5(0.8)^(n-1), so a = 5 and r = 0.8
  2. Check convergence condition Since |r| = |0.8| = 0.8 < 1, the series converges
  3. Apply the infinite geometric series formula Sum = a / (1 - r) Sum = 5 / (1 - 0.8)
  4. Calculate the denominator 1 - 0.8 = 0.2
  5. Divide to find the sum 5 / 0.2 = 25

The answer is 25.

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