AS & A Level · Practice questions · Convergence and the sum to infinity

Settle the sum, or show it runs away.

Six original Cambridge-style questions on convergence: limits of sums to \(n\) terms, sums to infinity by the method of differences, divergent series whose terms tend to zero and the least \(n\) that brings \(S_n\) within a tolerance of \(S\).

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Find \(S_n\), then let \(n \to \infty\).

Find \(S_n\) as a formula: standard results, partial fractions and differences, or the geometric formula.
Inspect every term containing \(n\): \(\dfrac{1}{n+k}\) and \(x^{n}\) with \(|x| < 1\) tend to 0; \(n\), \(\sqrt{n}\), \(\ln n\) grow.
Conclude in words: '... \(\to 0\) as \(n \to \infty\), so \(S_n \to S\)', or 'grows without limit, so divergent'.
Gap: \(S - S_n = \displaystyle\sum_{r=n+1}^{\infty} u_r\); solve the inequality for \(n\), round UP and check.
Never argue convergence from 'the terms tend to 0' alone.

Recall

[2 marks]

The sum of the first \(n\) terms of a series is \(S_n = \dfrac{n}{2n+1}\). Show that the series is convergent and state its sum to infinity.

Divide top and bottom by \(n\): \(S_n = \dfrac{1}{2 + \frac{1}{n}}\). ✓
As \(n \to \infty\), \(\dfrac{1}{n} \to 0\), so \(S_n \to \dfrac{1}{2}\): convergent with sum to infinity \(\dfrac{1}{2}\). ✓

Calculation

[5 marks]

(a) Express \(\dfrac{1}{(r+1)(r+3)}\) in partial fractions.

(b) Hence find \(\displaystyle\sum_{r=1}^{n} \frac{1}{(r+1)(r+3)}\) in terms of \(n\).

(a) \(\dfrac{1}{(r+1)(r+3)} = \dfrac{1}{2}\left(\dfrac{1}{r+1} - \dfrac{1}{r+3}\right)\). ✓
(b) The terms \(\tfrac{1}{4}, \tfrac{1}{5}, \dots, \tfrac{1}{n+1}\) cancel, leaving \(\dfrac{1}{2}\left(\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{n+2} - \dfrac{1}{n+3}\right)\). ✓
\(S_n = \dfrac{5}{12} - \dfrac{1}{2(n+2)} - \dfrac{1}{2(n+3)}\). ✓
(c) \(\dfrac{1}{2(n+2)} \to 0\) and \(\dfrac{1}{2(n+3)} \to 0\) as \(n \to \infty\). ✓
Sum to infinity \(= \dfrac{5}{12}\). ✓

Analysis

[4 marks]

(a) Show that \(\dfrac{1}{\sqrt{r} + \sqrt{r+1}} = \sqrt{r+1} - \sqrt{r}\).

(b) Hence find \(\displaystyle\sum_{r=1}^{n} \frac{1}{\sqrt{r} + \sqrt{r+1}}\) and determine whether the infinite series is convergent.

(a) Multiply top and bottom by \(\sqrt{r+1} - \sqrt{r}\): the denominator becomes \((r+1) - r = 1\). ✓
(b) By differences, \(S_n = \sqrt{n+1} - \sqrt{1} = \sqrt{n+1} - 1\). ✓
\(\sqrt{n+1} \to \infty\) as \(n \to \infty\), so \(S_n\) has no finite limit. ✓
The series is divergent, even though its terms tend to 0. ✓

Calculation

[5 marks]

(a) Show that \(\displaystyle\sum_{r=1}^{n} \frac{4}{(4r-1)(4r+3)} = \frac{1}{3} - \frac{1}{4n+3}\).

(b) Find the least value of \(n\) for which the sum to infinity exceeds \(S_n\) by less than \(10^{-4}\).

(a) \(\dfrac{4}{(4r-1)(4r+3)} = \dfrac{1}{4r-1} - \dfrac{1}{4r+3}\), since \((4r+3) - (4r-1) = 4\). ✓
Differences: \(\left(\tfrac{1}{3} - \tfrac{1}{7}\right) + \left(\tfrac{1}{7} - \tfrac{1}{11}\right) + \dots + \left(\dfrac{1}{4n-1} - \dfrac{1}{4n+3}\right) = \dfrac{1}{3} - \dfrac{1}{4n+3}\). ✓
(b) \(\dfrac{1}{4n+3} \to 0\), so \(S = \dfrac{1}{3}\) and \(S - S_n = \dfrac{1}{4n+3}\). ✓
\(\dfrac{1}{4n+3} < 10^{-4} \Rightarrow 4n + 3 > 10000 \Rightarrow n > 2499.25\). ✓
Least \(n = 2500\) (\(n = 2499\) gives \(\tfrac{1}{9999} > 10^{-4}\)). ✓

Proof

[5 marks]

(a) Show that \(\displaystyle\sum_{r=1}^{n} \ln\left(\frac{(r+1)^{2}}{r(r+2)}\right) = \ln\left(\frac{2(n+1)}{n+2}\right)\).

(b) Deduce that the infinite series converges and state its sum.

(c) Explain why \(\displaystyle\sum_{r=1}^{\infty} \ln\left(\frac{r+1}{r}\right)\) has no sum, even though its terms also tend to 0.

(a) Let \(f(r) = \ln\left(\dfrac{r+1}{r}\right)\). Then \(\ln\left(\dfrac{(r+1)^{2}}{r(r+2)}\right) = \ln\left(\dfrac{r+1}{r}\right) - \ln\left(\dfrac{r+2}{r+1}\right) = f(r) - f(r+1)\). ✓
Summing, everything cancels except \(f(1) - f(n+1) = \ln 2 - \ln\left(\dfrac{n+2}{n+1}\right) = \ln\left(\dfrac{2(n+1)}{n+2}\right)\). ✓
(b) \(\dfrac{2(n+1)}{n+2} = \dfrac{2 + \frac{2}{n}}{1 + \frac{2}{n}} \to 2\), so \(S_n \to \ln 2\): convergent, sum \(\ln 2\). ✓
(c) Its terms are \(\ln(r+1) - \ln r\), so by differences its sum to \(n\) terms is \(\ln(n+1)\), which grows without limit. ✓
So it diverges: terms tending to 0 does not guarantee convergence. ✓

Analysis

[7 marks]

(a) Show that \(\dfrac{1}{r(r+1)(r+2)} = \dfrac{1}{2}\left(\dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)}\right)\).

(b) Hence find \(S_n = \displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)(r+2)}\) and the sum to infinity \(S\).

(a) \(\dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)} = \dfrac{(r+2) - r}{r(r+1)(r+2)} = \dfrac{2}{r(r+1)(r+2)}\); halve both sides. ✓
(b) By differences, \(S_n = \dfrac{1}{2}\left(\dfrac{1}{1 \times 2} - \dfrac{1}{(n+1)(n+2)}\right) = \dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2)}\). ✓
\(\dfrac{1}{2(n+1)(n+2)} \to 0\), so \(S = \dfrac{1}{4}\). ✓
(c) \(S - S_n = \dfrac{1}{2(n+1)(n+2)} < 10^{-4} \Rightarrow (n+1)(n+2) > 5000\). ✓
\(n = 69\): \(70 \times 71 = 4970\), not enough; \(n = 70\): \(71 \times 72 = 5112 > 5000\). ✓
Least \(n = 70\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.