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Sum to Infinity & Convergence

CSEC Additional Mathematics Essential Knowledge: The sum to infinity of a geometric series is a fascinating concept where we can find the sum of an infinite number of terms, provided the series converges. This concept connects algebra with limits and has practical applications in finance, physics, and computer science.

Key Concept: For a geometric series with first term \(a\) and common ratio \(r\), if \(|r| < 1\), the series converges and its sum to infinity is given by \(S_\infty = \frac{a}{1-r}\). If \(|r| \geq 1\), the series diverges (has no finite sum).

Part 1: Geometric Series Fundamentals

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Geometric Progression Review

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Geometric Series Formula

A geometric series has the form:

\[a + ar + ar^2 + ar^3 + \cdots + ar^{n-1}\]

where:

\(a\) = first term
\(r\) = common ratio (each term divided by the previous term)
\(n\) = number of terms

The sum of the first \(n\) terms is:

\[S_n = \frac{a(1 – r^n)}{1 – r} \quad \text{for } r \neq 1\]

∞

The Infinite Series Concept

What happens when \(n \to \infty\)? Consider these two scenarios:

When \(|r| < 1\)

As \(n\) increases, \(r^n\) gets smaller and smaller, approaching 0:

\(r^n \to 0 \quad \text{as} \quad n \to \infty\)

Example: \(0.5^{10} = 0.000976\), \(0.5^{20} \approx 0.000000954\)

Result: The series converges to a finite sum

When \(|r| \geq 1\)

As \(n\) increases, \(r^n\) grows or oscillates:

\(r^n \to \infty \quad \text{or oscillates}\)

Examples: \(2^{10} = 1024\), \(2^{20} = 1,048,576\)

Result: The series diverges (no finite sum)

📝 Example 1: Identifying Convergence

Which of these geometric series converge?

(a) \(4 + 2 + 1 + 0.5 + \cdots\)

(b) \(3 + 6 + 12 + 24 + \cdots\)

For (a): First term \(a = 4\), common ratio \(r = \frac{2}{4} = 0.5\)

Check: \(|r| = |0.5| = 0.5 < 1\) ✓

Conclusion: Series (a) converges

For (b): First term \(a = 3\), common ratio \(r = \frac{6}{3} = 2\)

Check: \(|r| = |2| = 2 \geq 1\)

Conclusion: Series (b) diverges

Part 2: Sum to Infinity Formula

S∞

The Convergence Formula

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Derivation of \(S_\infty\)

Starting from the finite sum formula:

\[S_n = \frac{a(1 – r^n)}{1 – r}\]

When \(|r| < 1\), as \(n \to \infty\), \(r^n \to 0\). Therefore:

\[S_\infty = \lim_{n \to \infty} S_n = \frac{a(1 – 0)}{1 – r} = \frac{a}{1 – r}\]

Critical Condition: This formula only works when \(|r| < 1\). If \(|r| \geq 1\), \(S_\infty\) does not exist (the series diverges).

|r| < 1
Series Converges
\(S_\infty = \frac{a}{1-r}\)

|r| ≥ 1
Series Diverges
No Finite Sum

📝 Example 2: Basic Sum to Infinity

Find the sum to infinity of the geometric series: \(16 + 8 + 4 + 2 + \cdots\)

Identify values: \(a = 16\), \(r = \frac{8}{16} = 0.5\)

Check convergence: \(|r| = 0.5 < 1\) ✓ (series converges)

Apply formula: \(S_\infty = \frac{a}{1-r} = \frac{16}{1 – 0.5}\)

Calculate: \(= \frac{16}{0.5} = 32\)

Interpretation: The infinite sum \(16 + 8 + 4 + 2 + 1 + 0.5 + \cdots\) equals exactly 32

📝 Example 3: Fractional Common Ratio

Find the sum to infinity of: \(27 – 9 + 3 – 1 + \frac{1}{3} – \cdots\)

Identify values: \(a = 27\), \(r = \frac{-9}{27} = -\frac{1}{3}\)

Check convergence: \(|r| = |-\frac{1}{3}| = \frac{1}{3} < 1\) ✓

Apply formula: \(S_\infty = \frac{a}{1-r} = \frac{27}{1 – (-\frac{1}{3})}\)

Calculate: \(= \frac{27}{1 + \frac{1}{3}} = \frac{27}{\frac{4}{3}} = 27 \times \frac{3}{4} = \frac{81}{4} = 20.25\)

Note: Negative ratios alternate signs, but series still converges if \(|r| < 1\)

Part 3: Visualizing Convergence

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Graphical Understanding

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Partial Sums Approach Limit

Consider the series: \(10 + 5 + 2.5 + 1.25 + \cdots\) (\(a=10, r=0.5\))

Number of Terms (n)	Partial Sum (S_n)	Difference from S∞ = 20
1	10	10 away
2	15	5 away
3	17.5	2.5 away
4	18.75	1.25 away
5	19.375	0.625 away
10	19.980	0.020 away
∞	20	0 away

The partial sums get closer and closer to 20 but never exceed it. This is the concept of a limit.

Visualizing \(S_\infty = \frac{a}{1-r}\)

Imagine walking halfway to a wall each time:

Start at 0, wall at distance \(S\)
Step 1: Go \(\frac{1}{2}S\)
Step 2: Go \(\frac{1}{4}S\) (half of remaining distance)
Step 3: Go \(\frac{1}{8}S\)
… and so on forever

You approach the wall but (in theory) never quite touch it. The total distance covered = \(S\).

Part 4: Finding Unknown Values

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CSEC-Style Problems

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Common Problem Types

CSEC often asks:

Given \(S_\infty\) and \(a\), find \(r\)
Given \(S_\infty\) and \(r\), find \(a\)
Given a series converges, find range/values of \(r\)
Word problems involving infinite sums

📝 Example 4: Finding the Common Ratio

The sum to infinity of a geometric series is 45, and the first term is 15. Find the common ratio.

Use formula: \(S_\infty = \frac{a}{1-r}\)

Substitute known values: \(45 = \frac{15}{1-r}\)

Cross-multiply: \(45(1-r) = 15\)

Solve: \(45 – 45r = 15 \Rightarrow -45r = 15 – 45 = -30\)

Final step: \(r = \frac{-30}{-45} = \frac{2}{3}\)

Check convergence: \(|r| = \frac{2}{3} < 1\) ✓

📝 Example 5: Past Paper Question (CSEC May 2018)

The first term of a geometric progression is 80 and the sum to infinity is 100.

(a) Calculate the common ratio.

(b) Find the sum of the first 4 terms.

Part (a): Use \(S_\infty = \frac{a}{1-r}\)

Substitute: \(100 = \frac{80}{1-r} \Rightarrow 100(1-r) = 80\)

Solve for r: \(100 – 100r = 80 \Rightarrow -100r = -20 \Rightarrow r = 0.2\)

Part (b): Use \(S_n = \frac{a(1-r^n)}{1-r}\) with \(n=4\)

Calculate: \(S_4 = \frac{80(1 – 0.2^4)}{1 – 0.2} = \frac{80(1 – 0.0016)}{0.8}\)

Final answer: \(= \frac{80 \times 0.9984}{0.8} = 100 \times 0.9984 = 99.84\)

Part 5: Applications & Real-World Examples

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Practical Uses of Sum to Infinity

Real-World Applications:

Physics: Calculating total distance traveled by a bouncing ball that loses energy each bounce

Finance: Calculating the present value of perpetual annuities or consols

Computer Science: Analyzing algorithms with decreasing recursive calls

Economics: The multiplier effect in macroeconomics

Engineering: Signal processing and feedback systems

📝 Example 6: The Bouncing Ball Problem

A ball is dropped from a height of 10 meters. Each bounce reaches 60% of the previous height. What is the total distance traveled by the ball before it comes to rest?

Initial drop: 10 meters

First bounce: Up and down = \(2 \times (10 \times 0.6) = 12\) meters

Second bounce: \(2 \times (10 \times 0.6^2) = 2 \times 3.6 = 7.2\) meters

Pattern: Total distance = \(10 + 12 + 7.2 + 4.32 + \cdots\)

Identify series: After initial 10m, we have a geometric series: \(12 + 7.2 + 4.32 + \cdots\) with \(a=12\), \(r=0.6\)

Sum to infinity: \(S_\infty = \frac{12}{1-0.6} = \frac{12}{0.4} = 30\) meters

Total distance: \(10 + 30 = 40\) meters

Part 6: Common Mistakes & Exam Tips

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Avoiding Pitfalls

Common Mistakes to Avoid: 1. Using \(S_\infty\) formula when \(|r| \geq 1\) (series diverges!)
2. Forgetting to check the convergence condition first
3. Mishandling negative common ratios (remember \(|r| < 1\) not \(r < 1\))
4. Confusing \(S_n\) formula with \(S_\infty\) formula
5. Not recognizing a geometric series in word problems
6. Incorrectly identifying \(a\) (first term) in the series

Exam Strategy: When solving sum to infinity problems: (1) Identify it’s a geometric series, (2) Find \(a\) and \(r\), (3) Check \(|r| < 1\) (if not, series diverges), (4) Apply \(S_\infty = \frac{a}{1-r}\), (5) For word problems, set up the series carefully, considering if the first term is handled separately.

🎯 Key Concepts Summary

Convergence Condition: \(|r| < 1\) (absolute value of r less than 1)
Sum to Infinity Formula: \(S_\infty = \frac{a}{1-r}\)
Divergence: If \(|r| \geq 1\), no finite sum exists
Partial Sums: \(S_n = \frac{a(1-r^n)}{1-r}\) for finite n
Special Cases:
- If \(r\) is positive and \(< 1\): Terms decrease toward zero
- If \(r\) is negative and \(|r| < 1\): Terms oscillate but decrease in magnitude
Common CSEC Questions:
- Find \(S_\infty\) given \(a\) and \(r\)
- Find \(r\) given \(S_\infty\) and \(a\)
- Word problems with bouncing, investments, etc.
- Combine with finding specific terms of the GP

Comparison Table: Convergence vs Divergence

Common Ratio (r)	Condition	Convergence	Sum to Infinity	Example Series
\(0 < r < 1\)	\(\|r\| < 1\)	Converges	\(S_\infty = \frac{a}{1-r}\)	\(8 + 4 + 2 + 1 + \cdots\) (r=0.5)
\(-1 < r < 0\)	\(\|r\| < 1\)	Converges	\(S_\infty = \frac{a}{1-r}\)	\(9 – 3 + 1 – \frac{1}{3} + \cdots\) (r=-1/3)
\(r = 0\)	\(\|r\| < 1\)	Converges	\(S_\infty = a\)	\(5 + 0 + 0 + 0 + \cdots\)
\(r \geq 1\)	\(\|r\| \geq 1\)	Diverges	No finite sum	\(2 + 4 + 8 + 16 + \cdots\) (r=2)
\(r \leq -1\)	\(\|r\| \geq 1\)	Diverges	No finite sum	\(3 – 6 + 12 – 24 + \cdots\) (r=-2)

Quiz: Test Your Understanding

Sum to Infinity & Convergence Quiz

Question 1: Find the sum to infinity of the geometric series: \(12 + 6 + 3 + 1.5 + \cdots\)

Answer:
\(a = 12\), \(r = \frac{6}{12} = 0.5\)
\(|r| = 0.5 < 1\), so series converges
\(S_\infty = \frac{a}{1-r} = \frac{12}{1-0.5} = \frac{12}{0.5} = 24\)
The sum to infinity is 24.

Question 2: The sum to infinity of a geometric progression is 60 and the first term is 20. Find the common ratio.

Answer:
Using \(S_\infty = \frac{a}{1-r}\):
\(60 = \frac{20}{1-r} \Rightarrow 60(1-r) = 20\)
\(60 – 60r = 20 \Rightarrow -60r = -40\)
\(r = \frac{-40}{-60} = \frac{2}{3}\)
Check: \(|r| = \frac{2}{3} < 1\) ✓

Question 3: Determine whether the series \(1 – \frac{2}{3} + \frac{4}{9} – \frac{8}{27} + \cdots\) converges. If it does, find its sum to infinity.

Answer:
\(a = 1\), \(r = \frac{-2/3}{1} = -\frac{2}{3}\)
\(|r| = |-\frac{2}{3}| = \frac{2}{3} < 1\), so series converges
\(S_\infty = \frac{a}{1-r} = \frac{1}{1 – (-\frac{2}{3})} = \frac{1}{1 + \frac{2}{3}} = \frac{1}{\frac{5}{3}} = \frac{3}{5}\)
The sum to infinity is \(\frac{3}{5}\) or 0.6.

Question 4: For what values of \(x\) does the geometric series \(1 + (2x-1) + (2x-1)^2 + (2x-1)^3 + \cdots\) converge?

Answer:
Common ratio \(r = 2x-1\)
For convergence: \(|r| < 1\)
\(|2x-1| < 1\)
\(-1 < 2x-1 < 1\)
Add 1: \(0 < 2x < 2\)
Divide by 2: \(0 < x < 1\)
The series converges for \(0 < x < 1\).

Question 5: A geometric series has first term 5 and sum to infinity 7.5. Find the third term of the series.

Answer:
Step 1: Find \(r\) using \(S_\infty = \frac{a}{1-r}\)
\(7.5 = \frac{5}{1-r} \Rightarrow 7.5(1-r) = 5\)
\(7.5 – 7.5r = 5 \Rightarrow -7.5r = -2.5\)
\(r = \frac{2.5}{7.5} = \frac{1}{3}\)
Step 2: Third term = \(ar^2 = 5 \times (\frac{1}{3})^2 = 5 \times \frac{1}{9} = \frac{5}{9}\)

CSEC Exam Strategy: When answering sum to infinity questions: (1) Always state/write the condition for convergence (\(|r| < 1\)), (2) Show the formula \(S_\infty = \frac{a}{1-r}\), (3) Substitute values carefully, (4) Check your answer makes sense (e.g., if \(a\) is positive and \(0 a\)). Remember: Divergent series have NO finite sum – don’t try to apply the formula!