Home / A-Level Further Maths 9231 / Summation of series / Convergence and the sum to infinity
Cambridge International AS & A Level · Topic 1.3
A-Level Further Maths 9231 / Topic 1.3 / AS & A Level

Add up forever and still get a number.

Once you have a formula for \(S_n\), the sum of the first \(n\) terms, you can ask what happens as \(n\) grows without limit. If \(S_n\) settles on a finite value the series converges and that value is its sum to infinity; if \(S_n\) keeps growing it diverges. You decide by looking hard at the terms in \(n\) that are left in \(S_n\).

Topic 1.3All summation of series lessons →
The key idea

A series \(\displaystyle\sum_{r=1}^{\infty} u_r\) is convergent if \(S_n = \displaystyle\sum_{r=1}^{n} u_r\) tends to a finite limit \(S\) as \(n \to \infty\); then \(S\) is the sum to infinity. Find \(S_n\) first (standard results, differences, geometric formula), then look at each term containing \(n\): terms like \(\dfrac{1}{n+1}\) or \(\left(\tfrac{1}{2}\right)^{n}\) tend to 0, while terms like \(n^{2}\), \(\sqrt{n+1}\) or \(\ln(n+1)\) grow without limit and make the series divergent. Terms \(u_r \to 0\) is NOT enough for convergence.

n Sn 2 4 6 8 10 12 0.25 0.5 0.75 0 limit 3/4 S∞ − S₄ S₁ = 1/3 Sₙ = Σ 1/(r(r + 2)) from r = 1 to n
Fig. 1: the partial sums S_n of the series with terms 1/(r(r + 2)), plotted for n = 1 to 12. They rise towards the dashed line at 3/4 but never reach it; the blue arrow shows the gap S - S_n at n = 4
Section 01

What converges means.

A series has infinitely many terms, so you cannot add them all directly. Instead, find \(S_n\), the sum of the first \(n\) terms, as a formula in \(n\). If \(S_n\) gets as close as you like to a fixed number \(S\) as \(n \to \infty\), the series is convergent and you write \(\displaystyle\sum_{r=1}^{\infty} u_r = S\). The syllabus asks you to decide this by direct consideration of \(S_n\): look at every term that still contains \(n\) and say what it does.

Worked example Differences, then the limit
Given that \(\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+2)} = \frac{3}{4} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)}\), show that the series is convergent and state its sum to infinity.
  1. (The formula comes from \(\dfrac{1}{r(r+2)} = \dfrac{1}{2}\left(\dfrac{1}{r} - \dfrac{1}{r+2}\right)\) and the method of differences: \(\dfrac{1}{2}\left(1 + \dfrac{1}{2} - \dfrac{1}{n+1} - \dfrac{1}{n+2}\right)\).)
  2. As \(n \to \infty\), \(\dfrac{1}{2(n+1)} \to 0\) and \(\dfrac{1}{2(n+2)} \to 0\).
  3. So \(S_n \to \dfrac{3}{4} - 0 - 0\), a finite limit: the series converges (Fig. 1).
Answer: Convergent, \(\displaystyle\sum_{r=1}^{\infty} \frac{1}{r(r+2)} = \frac{3}{4}\)
Worked example A geometric check
Find \(\displaystyle\sum_{r=1}^{n} 3\left(\tfrac{1}{2}\right)^{r}\) and hence the sum to infinity.
  1. Geometric with first term \(a = \dfrac{3}{2}\) and ratio \(\dfrac{1}{2}\): \(S_n = \dfrac{\frac{3}{2}\left(1 - \left(\frac{1}{2}\right)^{n}\right)}{1 - \frac{1}{2}} = 3\left(1 - \left(\tfrac{1}{2}\right)^{n}\right)\).
  2. As \(n \to \infty\), \(\left(\tfrac{1}{2}\right)^{n} \to 0\), so \(S_n \to 3\).
  3. This agrees with \(S = \dfrac{a}{1 - x} = \dfrac{3/2}{1/2} = 3\): the familiar formula is just this limit.
Answer: \(S_n = 3 - 3\left(\tfrac{1}{2}\right)^{n}\), sum to infinity \(3\)
Examiner trap

Do not 'put \(n = \infty\)' into \(S_n\). Write the limit argument in words: '\(\dfrac{1}{n+1} \to 0\) as \(n \to \infty\), so \(S_n \to \dfrac{3}{4}\)'. The mark is for saying which terms tend to zero.

Section 02

When a series diverges.

If \(S_n\) grows without limit, or never settles, the series is divergent and has no sum to infinity. Series of positive terms whose \(S_n\) behaves like \(n\) or \(n^{2}\) are the obvious cases: \(\displaystyle\sum_{r=1}^{n} r = \tfrac{1}{2}n(n+1)\) is unbounded. The less obvious cases are series whose terms shrink to zero but whose sum still grows.

The harmonic series \(1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4} + \dots\) is the famous example: its terms tend to 0, yet its partial sums pass every number eventually (they grow roughly like \(\ln n\)). So 'the terms tend to zero' is necessary for convergence, but it is never enough on its own. Always find \(S_n\) and look at it.

Worked example Shrinking terms, growing sum
Find \(\displaystyle\sum_{r=1}^{n} \left(\sqrt{r+1} - \sqrt{r}\right)\) and decide whether the infinite series converges.
  1. Write out the terms: \((\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \dots + (\sqrt{n+1} - \sqrt{n})\).
  2. Everything cancels except \(\sqrt{n+1} - 1\).
  3. Each term \(\sqrt{r+1} - \sqrt{r} = \dfrac{1}{\sqrt{r+1} + \sqrt{r}}\) tends to 0, but \(S_n = \sqrt{n+1} - 1\) grows without limit as \(n \to \infty\).
Answer: \(S_n = \sqrt{n+1} - 1\); the series diverges
Worked example A logarithmic telescope
Show that \(\displaystyle\sum_{r=1}^{n} \ln\left(\frac{r+1}{r}\right) = \ln(n+1)\) and comment on convergence.
  1. \(\ln\left(\dfrac{r+1}{r}\right) = \ln(r+1) - \ln r\).
  2. Summing: \((\ln 2 - \ln 1) + (\ln 3 - \ln 2) + \dots + (\ln(n+1) - \ln n) = \ln(n+1) - \ln 1 = \ln(n+1)\).
  3. The terms \(\ln\left(1 + \dfrac{1}{r}\right) \to \ln 1 = 0\), but \(\ln(n+1) \to \infty\).
Answer: \(S_n = \ln(n+1)\), so the series is divergent even though its terms tend to 0
Your turn
Find \(\displaystyle\sum_{r=1}^{n} (2r + 1)\) and say whether the infinite series converges.
Examiner trap

'The terms tend to zero, so the series converges' is a classic wrong reason and scores nothing. Use the terms only to rule convergence OUT (if \(u_r \not\to 0\) it must diverge); to rule it IN, show that \(S_n\) has a finite limit.

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Four quick checks. Each correct answer earns XP and records this skill on your profile.

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A series has \(S_n = 5 - \dfrac{3}{n+2}\). What is its sum to infinity?

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Which of these sums to \(n\) terms belongs to a DIVERGENT series?

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The terms \(u_r\) of a series tend to 0 as \(r \to \infty\). What can you conclude?

Quick check+10 XP

Find \(\displaystyle\sum_{r=1}^{\infty} \left(\tfrac{1}{3}\right)^{r}\).

Section 03

The full exam chain.

A typical Paper 1 question runs in a chain: express \(u_r\) in partial fractions, find \(S_n\) by the method of differences, deduce the sum to infinity, then work with the gap \(S - S_n\), which is the sum of all the terms after the \(n\)th: \(S - S_n = \displaystyle\sum_{r=n+1}^{\infty} u_r\). For positive terms the gap is positive and shrinks to 0 (Fig. 1, blue arrow), so you can ask for the least \(n\) that makes \(S_n\) as accurate as you like.

Worked example Partial fractions to a tolerance
Find \(\displaystyle\sum_{r=1}^{n} \frac{2}{(2r-1)(2r+1)}\), deduce the sum to infinity, and find the least \(n\) for which \(S - S_n < 0.001\).
  1. Partial fractions: \(\dfrac{2}{(2r-1)(2r+1)} = \dfrac{1}{2r-1} - \dfrac{1}{2r+1}\).
  2. Differences: \(\left(1 - \tfrac{1}{3}\right) + \left(\tfrac{1}{3} - \tfrac{1}{5}\right) + \dots + \left(\dfrac{1}{2n-1} - \dfrac{1}{2n+1}\right) = 1 - \dfrac{1}{2n+1}\).
  3. \(\dfrac{1}{2n+1} \to 0\), so the sum to infinity is \(1\), and \(S - S_n = \dfrac{1}{2n+1}\).
  4. \(\dfrac{1}{2n+1} < 0.001 \Rightarrow 2n + 1 > 1000 \Rightarrow n > 499.5\).
  5. \(n\) is a whole number, so the least value is \(500\). (Check: \(n = 499\) gives \(\tfrac{1}{999} > 0.001\); \(n = 500\) gives \(\tfrac{1}{1001} < 0.001\).)
Answer: \(S_n = 1 - \dfrac{1}{2n+1}\), \(S = 1\), least \(n = 500\)
Worked example A gap that needs a quadratic
For \(\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+2)} = \frac{3}{4} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)}\), find the least \(n\) for which \(S_n\) is within \(0.01\) of the sum to infinity.
  1. \(S - S_n = \dfrac{1}{2(n+1)} + \dfrac{1}{2(n+2)} = \dfrac{2n+3}{2(n+1)(n+2)}\).
  2. Need \(\dfrac{2n+3}{2(n+1)(n+2)} < \dfrac{1}{100}\), so \(100(2n+3) < 2(n+1)(n+2)\) (both denominators positive).
  3. \(200n + 300 < 2n^{2} + 6n + 4 \Rightarrow n^{2} - 97n - 148 > 0\).
  4. The positive root is \(n = \dfrac{97 + \sqrt{10001}}{2} \approx 98.50\), so \(n > 98.50\).
  5. Check: \(n = 98\) gives \(\tfrac{199}{19800} \approx 0.01005\) (too big); \(n = 99\) gives \(\tfrac{201}{20200} \approx 0.00995\).
Answer: Least \(n = 99\)
Your turn
Using \(\displaystyle\sum_{r=1}^{n} \frac{2}{(2r-1)(2r+1)} = 1 - \frac{1}{2n+1}\), find \(\displaystyle\sum_{r=n+1}^{\infty} \frac{2}{(2r-1)(2r+1)}\) in terms of \(n\).
Examiner trap

When you solve for \(n\) you get a decimal such as \(n > 499.5\) or \(n > 98.50\). Round UP to the next whole number, and check the two whole numbers either side of the boundary; the boundary decimal itself is never the answer.

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Given \(\displaystyle\sum_{r=1}^{n} \frac{1}{(r+1)(r+2)} = \frac{1}{2} - \frac{1}{n+2}\), state the sum to infinity of the series.

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A series has \(S_n = 1 - \dfrac{1}{2n+1}\). Find the least value of \(n\) for which \(S - S_n < 0.01\), where \(S\) is the sum to infinity.

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Given that \(\displaystyle\sum_{r=1}^{n} \ln\left(\frac{r+1}{r}\right) = \ln(n+1)\), which statement is correct?

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Given \(\displaystyle\sum_{r=1}^{n} \frac{2}{(2r-1)(2r+1)} = 1 - \frac{1}{2n+1}\), find \(\displaystyle\sum_{r=n+1}^{\infty} \frac{2}{(2r-1)(2r+1)}\).

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Six original Cambridge-style questions
Limits of \(S_n\), sums to infinity after partial fractions, divergent series with shrinking terms, a logarithmic telescope and the least \(n\) for a given accuracy.