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Sequences and series

A sequence is an ordered list of numbers. A series is the result of adding terms of a sequence. That distinction is the starting point for the whole topic:

2,5,8,11,is a sequence,2,5,8,11,\ldots \qquad \text{is a sequence,}

whereas

2+5+8+11+is a series.2+5+8+11+\cdots \qquad \text{is a series.}

At A-level, you learn to describe terms, recognise arithmetic and geometric structure, calculate finite and infinite sums, use sigma notation, expand binomials and build models from repeated change.

You should be able to:

  • substitute into formulae and simplify expressions
  • solve linear and quadratic equations
  • work with powers, fractions and percentages
  • find an nth term from a simple pattern
  • use a scientific calculator accurately

Review foundation sequences, linear equations or the laws of indices if these skills are uncertain.

Write unu_n for the term in position nn. The subscript is a label, not multiplication: u5u_5 means the fifth term, not u×5u\times5.

A sequence can be given explicitly, with unu_n written directly in terms of nn:

un=3n1.u_n=3n-1.

Then u1=2u_1=2, u2=5u_2=5 and u20=59u_{20}=59.

Alternatively, a sequence can be given recursively, with each term defined from an earlier term:

u1=2,un+1=un+3.u_1=2,\qquad u_{n+1}=u_n+3.

The initial value is essential. The rule un+1=un+3u_{n+1}=u_n+3 describes infinitely many sequences until a starting term is supplied.

Worked example 1: generate a recursive sequence

Section titled “Worked example 1: generate a recursive sequence”

Given

u1=10,un+1=12un+3,u_1=10,\qquad u_{n+1}=\frac12u_n+3,

find u2u_2, u3u_3 and u4u_4.

Apply the rule one step at a time:

u2=12(10)+3=8,u3=12(8)+3=7,u4=12(7)+3=6.5.\begin{aligned} u_2&=\frac12(10)+3=8,\\ u_3&=\frac12(8)+3=7,\\ u_4&=\frac12(7)+3=6.5. \end{aligned}

Do not substitute n=2n=2 into the right side. It contains unu_n, not nn. Study sequences and recurrence relations for iteration, increasing and decreasing sequences, and long-term behaviour.

2. Arithmetic means add; geometric means multiply

Section titled “2. Arithmetic means add; geometric means multiply”

An arithmetic sequence has a constant difference dd:

un+1un=d.u_{n+1}-u_n=d.

If its first term is aa, then

un=a+(n1)d.\boxed{u_n=a+(n-1)d}.

A geometric sequence has a constant ratio rr:

un+1un=r,\frac{u_{n+1}}{u_n}=r,

where the ratio is defined. If its first term is aa, then

un=arn1.\boxed{u_n=ar^{n-1}}.

The exponent or multiple is n1n-1 because no step has occurred at the first term.

SequenceTestType
4,9,14,19,4,9,14,19,\ldotsdifferences are 55arithmetic
80,40,20,10,80,40,20,10,\ldotsratios are 12\frac12geometric
1,4,9,16,1,4,9,16,\ldotsneither is constantneither

An arithmetic sequence has u7=23u_7=23 and u18=67u_{18}=67. Find its first term and common difference.

Using un=a+(n1)du_n=a+(n-1)d:

a+6d=23,a+17d=67.a+6d=23,\qquad a+17d=67.

Subtracting gives

11d=44,11d=44,

so d=4d=4. Then a+24=23a+24=23, giving a=1a=-1.

The sequence is therefore

1,3,7,11,-1,3,7,11,\ldots

and un=1+4(n1)=4n5u_n=-1+4(n-1)=4n-5. Continue with arithmetic sequences and series.

Worked example 3: distinguish a term from a sum

Section titled “Worked example 3: distinguish a term from a sum”

For the geometric sequence with a=3a=3 and r=2r=2, find the sixth term and the sum of the first six terms.

The term is

u6=3(25)=96.u_6=3(2^5)=96.

The finite geometric sum is

S6=a(r61)r1=3(261)21=189.\begin{aligned} S_6&=\frac{a(r^6-1)}{r-1}\\ &=\frac{3(2^6-1)}{2-1}\\ &=189. \end{aligned}

Thus u6=96u_6=96 but S6=189S_6=189. Confusing unu_n with SnS_n is one of the most common errors in this topic. Learn finite sums, infinite sums and convergence in geometric sequences and series.

3. Infinite does not automatically mean infinite in value

Section titled “3. Infinite does not automatically mean infinite in value”

For a geometric series,

a+ar+ar2+,a+ar+ar^2+\cdots,

the terms approach zero when r<1|r|<1. Its partial sums then approach the finite limit

S=a1r,r<1.\boxed{S_\infty=\frac{a}{1-r}},\qquad |r|<1.

For example,

12+6+3+32+12+6+3+\frac32+\cdots

has a=12a=12 and r=12r=\frac12, so

S=1211/2=24.S_\infty=\frac{12}{1-1/2}=24.

The condition r<1|r|<1 is not optional. If r=2r=2, the terms grow. If r=1r=-1, they oscillate. In neither case do the partial sums approach one fixed number.

Worked example 4: form an infinite series from a recurring decimal

Section titled “Worked example 4: form an infinite series from a recurring decimal”

Express 0.2727270.272727\ldots as a fraction.

Separate the repeated blocks:

0.272727=0.27+0.0027+0.000027+.0.272727\ldots=0.27+0.0027+0.000027+\cdots.

This is geometric with

a=27100,r=1100.a=\frac{27}{100},\qquad r=\frac1{100}.

Therefore

S=27/10011/100=2799=311.S_\infty =\frac{27/100}{1-1/100} =\frac{27}{99} =\frac3{11}.

This works because r<1|r|<1.

The Greek capital letter sigma, Σ\Sigma, means “sum”. For example,

k=15(2k+1)\sum_{k=1}^{5}(2k+1)

means substitute k=1,2,3,4,5k=1,2,3,4,5 and add:

3+5+7+9+11=35.3+5+7+9+11=35.

The lower value gives the starting index and the upper value gives the final index. The number of terms is

51+1=5.5-1+1=5.

The index letter is temporary, so

k=15(2k+1)=j=15(2j+1).\sum_{k=1}^{5}(2k+1)=\sum_{j=1}^{5}(2j+1).

Study sigma notation before using standard sum formulae or manipulating indexed expressions.

For a non-negative integer nn, the binomial theorem states

(a+b)n=r=0n(nr)anrbr.(a+b)^n=\sum_{r=0}^{n}\binom nr a^{n-r}b^r.

The coefficient

(nr)=n!r!(nr)!\binom nr=\frac{n!}{r!(n-r)!}

counts the ways to choose which rr of the nn brackets contribute a factor of bb.

Worked example 5: find one term without expanding everything

Section titled “Worked example 5: find one term without expanding everything”

Find the coefficient of x3x^3 in (2x)6(2-x)^6.

The general term is

Tr+1=(6r)26r(x)r.T_{r+1}=\binom6r2^{6-r}(-x)^r.

An x3x^3 term requires r=3r=3. Its coefficient is

(63)23(1)3=208(1)=160.\binom632^3(-1)^3=20\cdot8\cdot(-1)=-160.

Notice that r=3r=3 gives the fourth term because the first term corresponds to r=0r=0. Learn this method in the binomial expansion for positive integer powers, then extend it to rational powers, where convergence restrictions become important.

Follow this order if you are learning the topic for the first time.

  1. Sequences and recurrence relations develops explicit and recursive definitions, iteration and sequence behaviour.
  2. Arithmetic sequences and series covers nth terms, finite sums and equations involving unknown terms.
  3. Geometric sequences and series covers constant ratios, finite sums, convergent infinite series and applications.
  4. Sigma notation expresses and manipulates sums compactly.
  5. The binomial expansion for positive integer powers uses combinations, Pascal’s triangle and the general term.
  6. The binomial expansion for rational powers introduces infinite expansions, approximations and validity intervals.
  7. Sequences and series in modelling applies repeated additive or multiplicative change and tests the assumptions of a model.

Given un=n23nu_n=n^2-3n, find u5u_5 and un+1u_{n+1}.

Answer

u5=523(5)=10.u_5=5^2-3(5)=10.

Replace every nn by n+1n+1:

un+1=(n+1)23(n+1)=n2n2.\begin{aligned} u_{n+1}&=(n+1)^2-3(n+1)\\ &=n^2-n-2. \end{aligned}

The expression un+1u_{n+1} does not mean un+1u_n+1.

Classify 162,54,18,6,162,54,18,6,\ldots and find its eighth term.

Answer

Each term is multiplied by 13\frac13, so the sequence is geometric with a=162a=162 and r=13r=\frac13.

u8=162(13)7=227.u_8=162\left(\frac13\right)^7=\frac2{27}.

Find 5+9+13++815+9+13+\cdots+81.

Answer

First count the terms:

5+(n1)4=81,5+(n-1)4=81,

so n=20n=20. Then

S20=202(5+81)=860.S_{20}=\frac{20}{2}(5+81)=860.

The last term alone does not tell you the number of terms.

Find the sum to infinity of 72.1+0.637-2.1+0.63-\cdots.

Answer

The ratio is r=0.3r=-0.3. Since r=0.3<1|r|=0.3<1, the series converges:

S=71(0.3)=7013.S_\infty=\frac7{1-(-0.3)}=\frac{70}{13}.

Use 1r1-r, including the sign of rr.

Evaluate

r=26(3r1).\sum_{r=2}^{6}(3r-1).

Answer

There are 62+1=56-2+1=5 terms:

5+8+11+14+17=55.5+8+11+14+17=55.

  • A sequence is not a series. unu_n denotes one term; SnS_n usually denotes a sum.
  • The nth term is not found by extending a pattern repeatedly. Use a formula when nn is large.
  • Arithmetic and geometric describe different operations. Arithmetic sequences add a constant; geometric sequences multiply by a constant.
  • An infinite sum needs a convergence check. Use S=a1rS_\infty=\frac{a}{1-r} only when r<1|r|<1.
  • Indices must be counted inclusively. From r=pr=p to r=qr=q there are qp+1q-p+1 terms.
  • A model has a domain. A geometric population model may predict non-integer values or unrealistic indefinite growth. Interpret the mathematics in context.

You are secure when you can move between a list, an nth-term formula, a recurrence relation, sigma notation and a contextual model without treating them as unrelated techniques. At the highest level, you should also justify convergence, choose an efficient formula rather than merely recall one, and check whether an algebraically valid answer is meaningful in its model.