What is convergence of a geometric series?

The partial sums of a geometric series form their own sequence S_n = a(1 - r^n)1 - r. This sequence converges exactly when r^n → 0, which happens if and only if |r| < 1. In that case

Find the limit of u_n = 2n^2 + 1n^2 + 3 as n → ∞. [2 marks]

SingaporeMathsSyllabus dot point

What does it mean for a series to converge, and how do we reason about the behaviour of a sequence as n grows?

Describe the behaviour of a sequence as n tends to infinity, determine the convergence of a geometric series, and interpret the limit of a sequence or partial sum

A focused answer to the H2 Mathematics outcome on convergence. The behaviour of a sequence as n tends to infinity, the convergence condition for a geometric series, and interpreting limits of sequences and partial sums.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to describe how a sequence behaves as $n$ grows without bound, to state and apply the convergence condition for a geometric series, and to interpret the limit of a sequence or of a sequence of partial sums. This formalises the intuition behind the sum to infinity.

The answer

Behaviour of a sequence as n tends to infinity

A sequence $u_n$ may converge to a limit $L$ (the terms settle ever closer to $L$ ), diverge to infinity, or oscillate. To find a limit of a rational expression in $n$ , divide numerator and denominator by the highest power of $n$ and use the fact that $\dfrac{1}{n^k} \to 0$ .

Convergence of a geometric series

The partial sums of a geometric series form their own sequence $S_n = \dfrac{a(1 - r^n)}{1 - r}$ . This sequence converges exactly when $r^n \to 0$ , which happens if and only if $|r| < 1$ . In that case

\lim_{n \to \infty} S_n = \frac{a}{1 - r}.

If $|r| \geq 1$ the partial sums grow without limit or oscillate, so the series diverges.

Convergence versus the terms tending to zero

A necessary condition for a series to converge is that its terms tend to zero. For a geometric series this is also sufficient, but in general "terms tend to zero" does not guarantee convergence. For the geometric case, $u_n = ar^{n-1} \to 0 \iff |r| < 1$ , neatly matching the series condition.

Interpreting a limit

A limit is the value the sequence (or partial sum) approaches but may never exactly reach. Saying a savings total "tends to" a figure means it gets arbitrarily close as time goes on, which is exactly the sum-to-infinity interpretation.

Worked example

A geometric series has first term $a$ and common ratio $r$ . Given that the sum to infinity is $18$ and the sum to infinity of the squares of the terms is $54$ , find $a$ and $r$ .

Step 1: Write the two sums to infinity

The terms are $a, ar, ar^2, \ldots$ , so $S_\infty = \dfrac{a}{1 - r} = 18$ .

The squares are $a^2, a^2 r^2, a^2 r^4, \ldots$ , a GP with first term $a^2$ and ratio $r^2$ , so $\dfrac{a^2}{1 - r^2} = 54$ .

Step 2: Use the difference of two squares

$1 - r^2 = (1 - r)(1 + r)$ , so $\dfrac{a^2}{(1-r)(1+r)} = 54$ .

Step 3: Divide the squared sum by the square of the first sum

$\dfrac{a^2 / [(1-r)(1+r)]}{[a/(1-r)]^2} = \dfrac{a^2 (1-r)^2}{(1-r)(1+r)\, a^2} = \dfrac{1 - r}{1 + r} = \dfrac{54}{18^2}$ ? Instead, compute $\dfrac{54}{18} = 3$ , where $\dfrac{a^2/(1-r^2)}{a/(1-r)} = \dfrac{a}{1+r} = 3$ .

Step 4: Solve the simultaneous equations

From $\dfrac{a}{1 - r} = 18$ and $\dfrac{a}{1 + r} = 3$ : dividing gives $\dfrac{1 + r}{1 - r} = 6$ , so $1 + r = 6 - 6r$ , giving $7r = 5$ , $r = \dfrac{5}{7}$ . Then $a = 18(1 - \tfrac{5}{7}) = 18 \times \tfrac{2}{7} = \dfrac{36}{7}$ .

Step 5: Check convergence

$|r| = \dfrac{5}{7} < 1$ , so both series converge, confirming the solution.

Common traps

Stating the wrong convergence condition: A geometric series converges if and only if $|r| < 1$ , not $r < 1$ (which would wrongly include $r \leq -1$ ).
Assuming terms-to-zero implies convergence in general: It is necessary but not sufficient for general series; only for the geometric case is it equivalent to $|r| < 1$ .
Forgetting to divide by the highest power: To find $\lim u_n$ for a rational sequence, divide top and bottom by the highest power of $n$ first.
Confusing a sequence with its series: The sequence is the list of terms; the series is the running total. Their convergence behaviours are related but distinct.
Treating a limit as attained: A convergent sequence approaches its limit; it need not equal it for any finite $n$ .

Examples in context

Example 1. A repeating decimal as a series. The decimal $0.\overline{3} = 0.3 + 0.03 + 0.003 + \cdots$ is geometric with $a = 0.3$ , $r = 0.1$ . Since $|r| < 1$ it converges to $\dfrac{0.3}{1 - 0.1} = \dfrac{1}{3}$ , explaining why $0.\overline{3} = \frac{1}{3}$ .

Example 2. Long-run dosage. A drug taken repeatedly, with a fraction $r = 0.7$ of the previous dose remaining each interval, builds toward a steady level $\dfrac{a}{1 - 0.7}$ . Because $|r| < 1$ the body burden converges rather than growing without bound.

Try this

Q1. Find the limit of $u_n = \dfrac{2n^2 + 1}{n^2 + 3}$ as $n \to \infty$ . [2 marks]

Cue. Divide by $n^2$ : $\dfrac{2 + 1/n^2}{1 + 3/n^2} \to 2$ .

Q2. State the condition for a geometric series to converge and explain why. [2 marks]

Cue. $|r| < 1$ , because then $r^n \to 0$ , so the partial sums tend to the finite limit $\dfrac{a}{1 - r}$ .

Q3. Determine whether the series with $a = 5$ , $r = 1.2$ converges. [1 mark]

Cue. $|r| = 1.2 \geq 1$ , so it diverges; no sum to infinity exists.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksA sequence is defined by

u_n = \dfrac{3n + 1}{n + 2}

. Describe its behaviour as

n \to \infty

and state its limit.

Show worked answer →

Divide numerator and denominator by $n$ : $u_n = \dfrac{3 + \frac{1}{n}}{1 + \frac{2}{n}}$ .

As $n \to \infty$ , $\dfrac{1}{n} \to 0$ and $\dfrac{2}{n} \to 0$ , so $u_n \to \dfrac{3}{1} = 3$ .

The sequence converges to the limit $3$ , approaching it from below (since for finite $n$ the numerator grows slightly slower relative to the denominator, giving values just under $3$ ).

Markers reward dividing by the highest power of $n$ , taking the limit of each term, and stating convergence to $3$ .

Original4 marksDetermine, with justification, whether the geometric series with first term

8

and common ratio

-\dfrac{3}{4}

converges, and if so find its sum to infinity.