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IB Mathematics AI SL The sum of infinite geometric sequences Study Notes - New Syllabus

IB Mathematics AI SL The sum of infinite geometric sequences Study Notes

LEARNING OBJECTIVE

Key Concepts: 

MAI HL and SL Notes – All topics

THE SUM OF ∞ TERMS IN A G.S. 

♦ Convergence Condition: \( |r| < 1 \)
\( S_\infty = \frac{u_1}{1 – r} \)

Example

\( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots =? \)

▶️Answer/Explanation

Solution:

$1$

Example

(i) \( 0.333… =? \) 
 (ii) \( 0.999… = ? \) 

▶️Answer/Explanation

Solution:

(i) $\frac{1}{3}$ (proof via G.S.)
(ii) $1$ (multiple proofs).

♦ Proof of \( S_\infty \)
1. Algebraic subtraction:
\( S_\infty = u_1 + r S_\infty \) → \( S_\infty = \frac{u_1}{1 – r} \).

Proof 3 (Alternative Approach)
We can derive the infinite sum formula by observing that:
\( S_\infty = u_1 + u_1 r + u_1 r^2 + u_1 r^3 + \cdots \)
Factor out \( u_1 \):
\( S_\infty = u_1 (1 + r + r^2 + r^3 + \cdots) \)
The series in parentheses is a geometric series with first term 1 and ratio \( r \). Thus:
\( 1 + r + r^2 + r^3 + \cdots = \frac{1}{1 – r} \quad \text{(for } |r| < 1) \)
Therefore:
\( S_\infty = \frac{u_1}{1 – r} \)

Key Points:
 The series converges only when \( |r| < 1 \).
 If \( |r| \geq 1 \), the series diverges (i.e., the sum grows infinitely or oscillates).

Example (Application)

Consider the repeating decimal \( 0.\overline{12} = 0.121212\ldots \). Express it as a fraction using the infinite geometric series formula.

▶️Answer/Explanation

Solution:

Rewrite the decimal as:
\( 0.\overline{12} = 0.12 + 0.0012 + 0.000012 + \cdots \)
This is a geometric series with:
 \( u_1 = 0.12 \)
 \( r = 0.01 \)

Since \( |r| < 1 \), the sum converges:
\( S_\infty = \frac{u_1}{1 – r} = \frac{0.12}{1 – 0.01} = \frac{0.12}{0.99} = \frac{12}{99} = \frac{4}{33} \)

Verification:
Divide 4 by 33 to confirm: \( 4 \div 33 = 0.\overline{12} \).

Example(Divergent Case)

Show that the series \( 1 + 2 + 4 + 8 + \cdots \) diverges.

▶️Answer/Explanation

Solution:

This is a geometric series with \( u_1 = 1 \) and \( r = 2 \). Since \( |r| \geq 1 \), the series diverges. The partial sums grow infinitely:
\( S_n = 2^n – 1 \rightarrow \infty \quad \text{as } n \rightarrow \infty \)

♦ Final Notes:
1. Convergence Test: Always check \( |r| < 1 \) before applying \( S_\infty \).
2. Exact vs. Approximate: For \( r \) values like \( \frac{1}{2} \), the sum is exact. For irrational \( r \) (e.g., \( \frac{1}{\pi} \)), the sum is exact in fractional form but may be approximated numerically.
3. Applications: Infinite series are used in finance (perpetuities), physics (waveforms), and computer science (algorithms).

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