Year 12 Maths Advanced (2027) Sequences and series

Limiting Sum of a Geometric Series

20 practice questions 0 video lessons Theory + worked examples

NSW · Year 12 Mathematics Advanced · Sequences and series

The limiting sum of an infinite geometric series exists when \(|r|<1\), and equals \(S_\infty=\dfrac{a}{1-r}\).

Part of the NSW Year 12 Mathematics Advanced course, in the Sequences and series focus area of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Theory

The limiting sum of a geometric series is the total of infinitely many terms, which is finite only when \(|r|<1\). This Year 12 Mathematics Advanced topic (MAV-12-03) uses \(S_\infty=\dfrac{a}{1-r}\).

When \(|r|<1\) the terms shrink toward \(0\), so adding infinitely many gives a finite total — the limiting sum \(S_\infty=\dfrac{a}{1-r}\).

This works because \(r^{\,n}\to0\) as \(n\to\infty\), so \(S_n=\dfrac{a(1-r^{\,n})}{1-r}\to\dfrac{a}{1-r}\). If \(|r|\ge1\) the terms do not shrink and there is no limiting sum.

Partial sums of \(8+4+2+\cdots\) climb toward \(S_\infty=16\).

\[S_\infty=\dfrac{a}{1-r}\qquad(|r|<1)\]

S infinity equals a over (1 minus r) for mod r less than 1

Method

Check that \(|r|<1\) (otherwise there is no limiting sum).
Identify the first term \(a\) and ratio \(r\).
Substitute into \(S_\infty=\dfrac{a}{1-r}\).

Example 1 — Basic

Find the limiting sum of \(8+4+2+\cdots\)

Solution

\(S_\infty\)	\(=\)	\(\dfrac{8}{1-\tfrac12}\)
	\(=\)	\(16\)

Example 2 — Negative ratio

Find the limiting sum of \(12-6+3-\cdots\)

Solution

\(S_\infty\)	\(=\)	\(\dfrac{12}{1-(-\tfrac12)}\)
	\(=\)	\(8\)

Example 3 — Find \(a\)

A series has \(S_\infty=20\) and \(r=\dfrac34\). Find \(a\).

Solution

\(a\)	\(=\)	\(S_\infty(1-r)\)
	\(=\)	\(20\times\tfrac14=5\)

Example 4 — Converge?

For \(1+\tfrac13+\tfrac19+\cdots\): state \(r\), find \(S_\infty\), does it converge?

Solution

\(r\)	\(=\)	\(\tfrac13\)
\(S_\infty\)	\(=\)	\(\dfrac{1}{1-\tfrac13}=\tfrac32\)
\(\ \)		\(\text{yes, }\|r\|<1\)

Common pitfalls

Check the ratio first. A limiting sum exists only when \(-1

\(a\) is the first term, not a later one.

Negative ratios can still converge. \(|r|<1\) allows \(r=-\dfrac12\).

Frequently asked questions

What is the limiting sum of a geometric series?

It is the total of infinitely many terms, given by S infinity equals a divided by (1 minus r). It exists only when the common ratio r is between minus 1 and 1.

When does an infinite geometric series have a sum?

Only when the absolute value of the common ratio is less than 1, so the terms shrink toward zero. Otherwise the sum grows without bound.

How do you find the first term given the limiting sum?

Rearrange the formula to a equals S infinity times (1 minus r).

Does 8 plus 4 plus 2 and so on have a finite sum?

Yes. The ratio is one half, which is less than 1, so the limiting sum is 8 divided by (1 minus one half), which is 16.

← Previous subtopic

Geometric Series (Summing a GP)

Next subtopic →

Recurring Decimals as Infinite GPs