Summation of finite and infinite series
Use standard power sums, telescoping differences and partial sums to evaluate finite series and decide convergence.
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Use standard results for sums of r, r² and r³
- Evaluate finite sums by the method of differences
- Decide convergence from a sum to n terms and find a limit when it exists
Standard power sums and telescoping
For positive integers r, Σr=n(n+1)/2, Σr²=n(n+1)(2n+1)/6 and Σr³=[n(n+1)/2]². Substitute the upper limit only after matching the summation's starting value.
Rewrite each term as a difference of neighbouring expressions. In a sum such as Σ(1/r−1/(r+1)), the middle terms cancel in pairs, leaving the first term minus the final shifted term.
Partial fractions often expose the difference: 1/[r(r+1)]=1/r−1/(r+1). Write the first several terms once to check the cancellation pattern before generalising.
Convergence is a limit of partial sums
A series converges when its partial sum Sₙ approaches a finite value as n grows. A geometric series is one useful case, but the syllabus also expects direct examination of partial sums after telescoping.
For a finite sum, keep the upper limit visible in the final expression. To obtain a sum to infinity, state the limit and the condition that makes the omitted term tend to zero.
Pause and explain
Which habit best checks your work on summation of finite and infinite series?
Worked example
Evaluate Σ from r=1 to n of 1/[r(r+1)], then find the sum to infinity.
Show the worked solution
- Use 1/[r(r+1)]=1/r−1/(r+1).
- The finite sum is (1−1/2)+(1/2−1/3)+⋯+(1/n−1/(n+1))=1−1/(n+1).
- As n tends to infinity, 1/(n+1) tends to zero.
Answer Sₙ=n/(n+1), and the infinite series sums to 1.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Evaluate Σ from r=1 to n of 1/[r(r+2)].
Give me a hint
Split the term into partial fractions, then list the first and last surviving terms.
Compare my reasoning
- 1/[r(r+2)]=½(1/r−1/(r+2)).
- The first two positive terms are 1 and 1/2; the last two negative terms are 1/(n+1) and 1/(n+2).
- Multiply their difference by 1/2.
Sₙ=3/4−1/[2(n+1)]−1/[2(n+2)].
Look for these in your work
- Partial fractions correct
- Both leading terms retained
- Both trailing terms retained
Find the sum of Σ from r=1 to n of (3r²−r).
Give me a hint
Use Σr²=n(n+1)(2n+1)/6 and Σr=n(n+1)/2.
Compare my reasoning
- Separate the sums: 3Σr²−Σr.
- Substitute the standard formulas.
- Factor n(n+1): n(n+1)[(2n+1)/2−1/2].
n²(n+1).
Look for these in your work
- Power-sum formulas correct
- Coefficients applied
- Expression simplified
A partial sum is Sₙ=5−2/(n+1). Decide whether the corresponding series converges and state its sum.
Give me a hint
Take the limit of Sₙ as n grows; convergence is about the partial sums.
Compare my reasoning
- The term 2/(n+1) tends to 0.
- Therefore Sₙ tends to 5.
- The series converges to the limiting partial sum.
It converges, with sum 5.
Look for these in your work
- Limit term evaluated
- Convergence decision tied to Sₙ
- Limit stated
Common mistakes
- Applying a standard sum formula to a series with a different starting value without adjusting it
- Dropping a boundary term when telescoping
- Calling a series convergent without taking the limit of its partial sums
What can you explain now?
What remains from Σ(r=1 to n) (1/r−1/(r+1))?
Compare with the explanation
1−1/(n+1)
Every intermediate reciprocal appears once with each sign and cancels.
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Make it stick
Why does telescoping work, and what limit proves that an infinite series converges?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.