Infinite Series (Grade 12 NSC Matric Mathematics): Revision Notes

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Infinite Series

What is an infinite series?

An infinite series is the sum of an unlimited number of terms in a sequence. Unlike finite series where we only add the first n terms, infinite series continue indefinitely. The key question becomes: what happens when we add infinitely many terms together?

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Surprisingly, even when adding infinitely many numbers, the answer isn't always infinity. Sometimes these infinite sums approach a specific finite value, which opens up fascinating mathematical possibilities.

Understanding convergence through investigation

When we examine what happens as we add more and more terms to a series, we can observe an important pattern. Let's consider the geometric series with first term T1=12T_1 = \frac{1}{2} and common ratio r=12r = \frac{1}{2}.

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Investigation: Partial Sums of a Geometric Series

As we add successive terms, we notice that:

  • The partial sum SnS_n gets larger: 12<34<78<1516<...\frac{1}{2} < \frac{3}{4} < \frac{7}{8} < \frac{15}{16} < ...
  • The amount by which SnS_n increases gets smaller: 12>14>18>116>...\frac{1}{2} > \frac{1}{4} > \frac{1}{8} > \frac{1}{16} > ...
  • The partial sum appears to be approaching the value 1

This behaviour suggests that even though we're adding infinitely many terms, the sum approaches a finite limit.

Convergence and divergence

Key definitions

Understanding whether a series converges or diverges is fundamental to working with infinite series. The following definitions help us classify the behaviour of infinite series:

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Convergent series: If the sum of a series gets closer and closer to a certain value as we increase the number of terms, we say the series converges. There is a limit to the sum of a converging series.

Divergent series: If a series does not converge, we say it diverges. The sum of an infinite series usually tends to infinity, but there are special cases where it does not.

Notation: We express the sum of an infinite number of terms as:

S∞=∑i=1∞TiS_{\infty} = \sum_{i=1}^{\infty} T_i

Infinite geometric series

Test for convergence

For geometric series, there's a simple test to determine convergence or divergence based on the common ratio:

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Test for convergence:

  • If −1<r<1-1 < r < 1, then the infinite geometric series converges
  • If r<−1r < -1 or r>1r > 1, then the infinite geometric series diverges

This test is crucial because it tells us immediately whether an infinite geometric series will have a finite sum or not.

Sum to infinity formula

When a geometric series converges (when −1<r<1-1 < r < 1), we can calculate its exact sum using a specific formula:

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Sum to infinity formula:

S∞=a1−rwhere −1<r<1S_{\infty} = \frac{a}{1-r} \quad \text{where } -1 < r < 1

Where:

  • aa is the first term of the series
  • rr is the common ratio

Understanding why the formula works

The derivation of this formula comes from examining the behaviour of partial sums as nn approaches infinity. As nn approaches infinity in a convergent geometric series, rnr^n approaches 0. This means:

  • The formula Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r} becomes S∞=a(1−0)1−r=a1−rS_{\infty} = \frac{a(1-0)}{1-r} = \frac{a}{1-r}
  • The series reaches its limiting value

Worked examples

The following examples demonstrate how to apply the concepts of convergence and sum to infinity in different contexts:

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Worked Example 1: Finding sum to infinity

Question: Find the sum to infinity of the series 18+6+2+...18 + 6 + 2 + ...

Solution:

Step 1: Identify the first term and common ratio

  • a=18a = 18
  • r=618=13r = \frac{6}{18} = \frac{1}{3}

Step 2: Check convergence

Since −1<13<1-1 < \frac{1}{3} < 1, the series converges.

Step 3: Apply the formula

S∞=a1−r=181−13=1823=18×32=27S_{\infty} = \frac{a}{1-r} = \frac{18}{1-\frac{1}{3}} = \frac{18}{\frac{2}{3}} = 18 \times \frac{3}{2} = 27

Therefore, the sum approaches 27 as we add infinitely many terms.

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Worked Example 2: Converting recurring decimals

Question: Convert 0.5‾0.\overline{5} to a fraction using infinite series.

Solution:

Step 1: Express as a series

0.5‾=0.5+0.05+0.005+...=510+5100+51000+...0.\overline{5} = 0.5 + 0.05 + 0.005 + ... = \frac{5}{10} + \frac{5}{100} + \frac{5}{1000} + ...

Step 2: Identify the geometric series

  • First term: a=510a = \frac{5}{10}
  • Common ratio: r=110r = \frac{1}{10}

Step 3: Check convergence and apply formula

Since −1<110<1-1 < \frac{1}{10} < 1, the series converges.

S∞=a1−r=5101−110=12910=59S_{\infty} = \frac{a}{1-r} = \frac{\frac{5}{10}}{1-\frac{1}{10}} = \frac{\frac{1}{2}}{\frac{9}{10}} = \frac{5}{9}

Therefore, 0.5‾=590.\overline{5} = \frac{5}{9}.

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Worked Example 3: Finding unknown values

Question: If ∑n=1∞arn−1=5\sum_{n=1}^{\infty} ar^{n-1} = 5, find the possible values of aa and rr.

Solution:

Step 1: Apply the sum formula

S∞=a1−r=5S_{\infty} = \frac{a}{1-r} = 5

Therefore: a=5(1−r)a = 5(1-r)

Step 2: Apply convergence condition

For convergence: −1<r<1-1 < r < 1

Substituting a=5(1−r)a = 5(1-r): −1<r<1-1 < r < 1

This gives us: −1<5−a5<1-1 < \frac{5-a}{5} < 1

Simplifying: 0<a<100 < a < 10

Step 3: State the answer

For the series to converge: 0<a<100 < a < 10 and −1<r<1-1 < r < 1.

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Key Points to Remember:

  • Infinite series can have finite sums when they converge to a specific limit
  • Convergence test: A geometric series converges if and only if −1<r<1-1 < r < 1
  • Sum to infinity formula: S∞=a1−rS_{\infty} = \frac{a}{1-r} for convergent geometric series
  • Divergent series have no finite limit and usually tend to infinity
  • Recurring decimals can be converted to fractions using infinite geometric series
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Exam Tips:

  • Always check the convergence condition first before applying the sum to infinity formula
  • Arithmetic series never converge - they always tend to positive or negative infinity
  • When converting recurring decimals, identify the repeating pattern as a geometric series
  • Remember that ∣r∣<1|r| < 1 is the key condition for geometric series convergence
  • Practice identifying the first term and common ratio accurately

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