Sum to infinity of a geometric series for A-Level Maths
The sum to infinity of a geometric series is the value that the sum of all the terms approaches as you keep adding more and more of them. For a series with first term a and common ratio r, the sum to infinity is S = a / (1 – r), but only when |r| < 1. If |r| is greater than or equal to 1, the sum does not converge and the formula does not apply.
This guide covers the formula, why the convergence condition matters, how to spot a geometric series in an Edexcel A-Level question, and the most common ways students lose marks in Paper 1 and Paper 2.
One formula to memorise
S = a / (1 – r). The first term a sits on top, the common ratio r sits inside the denominator. It is on the Edexcel formula sheet but learn it anyway.
Convergence is non-negotiable
The series only converges when |r| < 1. State this condition in every answer or you lose a mark even if the arithmetic is correct.
Approaches but never quite reaches
Each term is smaller than the last, so the running total gets closer and closer to S without ever exceeding it. This is the formal idea of a limit.
What a geometric series actually is
A geometric series is a sum of terms where each term is found by multiplying the previous one by a fixed number called the common ratio, r. The general form is a + ar + ar² + ar³ + ... where a is the first term.
Examples of geometric series include 1 + 1/2 + 1/4 + 1/8 + ... (where a = 1 and r = 1/2) and 3 + 6 + 12 + 24 + ... (where a = 3 and r = 2). Only the first of these has a finite sum to infinity, because |r| < 1.
Spotting r from the terms Divide any term by the term before it. If the ratio is the same every time, you have a geometric series, and that ratio is r. For example, in 81 + 54 + 36 + 24 + ..., 54/81 = 2/3 and 36/54 = 2/3, so r = 2/3.
The sum to infinity formula
The sum of the first n terms of a geometric series is Sₙ = a(1 – rⁿ) / (1 – r). When |r| < 1, rⁿ tends to zero as n tends to infinity, so the formula simplifies to S = a / (1 – r).
When |r| is greater than or equal to 1, rⁿ does not tend to zero. Instead it grows without bound (if |r| > 1) or stays at 1 (if r = 1), and the sum has no finite limit. That is why the condition |r| < 1 is built into the formula.
| Quantity | Symbol | Formula | When to use |
|---|---|---|---|
| First term | a | Given in the question | Always identify a first |
| Common ratio | r | Any term divided by the previous one | Check |r| < 1 before using sum to infinity |
| Sum of first n terms | Sₙ | a(1 – rⁿ) / (1 – r) | Finite sums, any value of r ≠ 1 |
| Sum to infinity | S∞ | a / (1 – r) | Only when |r| < 1 |
Why the convergence condition matters
Edexcel mark schemes consistently award a separate mark for stating |r| < 1. If you find r, plug it into the formula, and write down an answer, you can still lose a mark for not justifying that the series converges in the first place.
This is true even when r is obviously less than 1. A line as short as the series converges because |r| = 1/3 < 1 is enough to pick up the mark. Get into the habit of stating it every time you use S = a / (1 – r).
What to write for the convergence mark For a series with r = 1/4, write: The series converges because |r| = 1/4 < 1, so the sum to infinity exists. That single sentence is worth one mark in nearly every question on the topic.
Worked example: Finding the sum to infinity
The first term of a geometric series is 12 and the third term is 3. Find the sum to infinity.
Step 1: Use the third term to find r. The third term is ar², so 12r² = 3, which gives r² = 1/4 and r = ±1/2.
Step 2: Check the convergence condition. Both possible values satisfy |r| < 1, so the sum to infinity exists for either.
Step 3: Apply S = a / (1 – r). If r = 1/2, S = 12 / (1 – 1/2) = 12 / (1/2) = 24. If r = –1/2, S = 12 / (1 – (–1/2)) = 12 / (3/2) = 8.
The two possible sums are 24 (for r = 1/2) and 8 (for r = –1/2). Edexcel usually wants both answers unless the question tells you the terms are all positive.
Worked example: Solving for the common ratio
A geometric series has first term 20 and sum to infinity 25. Find the common ratio.
Step 1: Substitute into S = a / (1 – r): 25 = 20 / (1 – r).
Step 2: Rearrange. 25(1 – r) = 20, so 1 – r = 20/25 = 4/5, which gives r = 1/5.
Step 3: Check convergence. |r| = 1/5 < 1, so the sum to infinity is valid.
The common ratio is 1/5. The first three terms of the series would be 20, 4, 0.8, and they would keep getting smaller until the running total approaches 25 from below.
Where students lose marks on sum to infinity questions
Edexcel examiner reports highlight the same recurring slips year after year. Most of them are small but cost real marks across a long-answer question.
Common mistakes that cost easy marks Forgetting to state the convergence condition |r| < 1. Using S = a / (1 – r) when |r| is greater than or equal to 1. Confusing the sum of the first n terms with the sum to infinity. Mixing up a (first term) with d (common difference, which belongs to arithmetic series). Dropping the negative when r is negative, for example writing 1 – (–1/2) = 1/2 instead of 3/2. Forgetting that ar² (not ar) gives the third term, because the first term is ar⁰ = a.
Key facts to memorise for the exam
- General term: The nth term of a geometric series is arⁿ⁻¹
- Common ratio: The ratio r equals any term divided by the previous term
- Sum of first n terms: Sₙ = a(1 – rⁿ) / (1 – r)
- Sum to infinity: S = a / (1 – r), only when |r| < 1
- Always state |r| < 1 before using the formula
- Two solutions: If r² appears, there may be a positive and a negative value of r
- Arithmetic vs geometric: Arithmetic uses a common difference d, geometric uses a common ratio r
- On the formula sheet: The sum to infinity is given, but you still need to know when to use it
