Study guides › 9231 FP1 › Summation of series Maths4U Original · Learn, check, explain
Converging series — 9231 FP1
Cancel interior terms before taking a limit.
1. Understand 2. See examples 3. Check answers 4. Explain your working Notes and key methods A difference such as \(\displaystyle \frac{1}{[r(r+ 1)]}= \frac{1}{r}- \frac{1}{r+ 1}\) telescopes when summed over consecutive integers. Write out the first and last few terms to see what survives.
Establish a finite partial-sum formula before letting n tend to infinity. An infinite series converges only if the remainder has a finite limiting value; termwise cancellation by itself is not a convergence proof.
Check the conditions Use finite partial sums before an infinite limit. Track the starting index and every uncancelled endpoint. Three worked examples Example 1 · Telescoping series and convergence For \(a=3,\ n=6\), consider \(\displaystyle S_n=\sum_{r=a}^{a+n-1}\frac{4}{r(r+1)}\).
Find \(\displaystyle S_{n}\). Find the sum from \(\displaystyle r=a\) to infinity. Find the positive remainder after n terms. Find the first term of the series. Show the reasoning (a) \(\displaystyle 4(\frac{1}{3}- \frac{1}{9})= \frac{8}{9}\)
Answer: \(\displaystyle \frac{8}{9}\).
(b) \(\displaystyle \frac{4}{a}= \frac{4}{3}\)
Answer: \(\displaystyle \frac{4}{3}\).
(c) \(\displaystyle \frac{4}{a+ n}= \frac{4}{9}\)
Answer: \(\displaystyle \frac{4}{9}\).
(d) \(\displaystyle \frac{4}{[3(4)]}= \frac{1}{3}\)
Answer: \(\displaystyle \frac{1}{3}\).
Example 2 · Telescoping series and convergence For \(a=4,\ n=7\), consider \(\displaystyle S_n=\sum_{r=a}^{a+n-1}\frac{2}{r(r+1)}\).
Find \(\displaystyle S_{n}\). Find the sum from \(\displaystyle r=a\) to infinity. Find the positive remainder after n terms. Find the first term of the series. Show the reasoning (a) \(\displaystyle 2(\frac{1}{4}- \frac{1}{11})= \frac{7}{22}\)
Answer: \(\displaystyle \frac{7}{22}\).
(b) \(\displaystyle \frac{2}{a}= \frac{1}{2}\)
Answer: \(\displaystyle \frac{1}{2}\).
(c) \(\displaystyle \frac{2}{a+ n}= \frac{2}{11}\)
Answer: \(\displaystyle \frac{2}{11}\).
(d) \(\displaystyle \frac{2}{[4(5)]}= \frac{1}{10}\)
Answer: \(\displaystyle \frac{1}{10}\).
Example 3 · Telescoping series and convergence For \(a=5,\ n=8\), consider \(\displaystyle S_n=\sum_{r=a}^{a+n-1}\frac{3}{r(r+1)}\).
Find \(\displaystyle S_{n}\). Find the sum from \(\displaystyle r=a\) to infinity. Find the positive remainder after n terms. Find the first term of the series. Show the reasoning (a) \(\displaystyle 3(\frac{1}{5}- \frac{1}{13})= \frac{24}{65}\)
Answer: \(\displaystyle \frac{24}{65}\).
(b) \(\displaystyle \frac{3}{a}= \frac{3}{5}\)
Answer: \(\displaystyle \frac{3}{5}\).
(c) \(\displaystyle \frac{3}{a+ n}= \frac{3}{13}\)
Answer: \(\displaystyle \frac{3}{13}\).
(d) \(\displaystyle \frac{3}{[5(6)]}= \frac{1}{10}\)
Answer: \(\displaystyle \frac{1}{10}\).
Common mistakes Do not cancel the first term against something outside the sum. A term tending to zero is necessary but not sufficient for convergence. Practise with answer checking Original exam-style problems Write complete solutions before opening the worked solution. The marking guidance rewards reasoning as well as answers.
Next skill: End-of-chapter review exercise 3 →
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