Num­bers often come in lists that fol­low a rule: the odd num­bers, the seats in the rows of a the­atre, the money in a sav­ings account month by month. Such a list is called a sequence, and adding up its terms gives a series. This les­son sets out the words and nota­tion used for the whole chap­ter, so that arith­metic, geo­met­ric and har­monic pro­gres­sions can be described pre­cisely in the lessons that fol­low.

Sequence

A set of num­bers arranged in a def­i­nite order accord­ing to some def­i­nite rule is called a sequence.

For exam­ple, 2,4,6,8,10,…2, 4, 6, 8, 10, \ldots is a sequence: the rule is "the even num­bers in increas­ing order". So is 1,4,9,16,25,…1, 4, 9, 16, 25, \ldots, the squares of the nat­ural num­bers.

The num­bers in a sequence are called its terms. We write the first term as t1t_1, the sec­ond as t2t_2, and so on. The term in posi­tion nn is tnt_n, called the nnth term or gen­eral term. (Some books write ana_n instead of tnt_n; the mean­ing is the same.)

The impor­tant thing is that the order mat­ters. The num­bers 1,3,51, 3, 5 and 5,3,15, 3, 1 are dif­fer­ent sequences, even though they con­tain the same num­bers.

Describ­ing a sequence by its gen­eral term

The neat­est way to give a sequence is a for­mula for tnt_n. Put n=1,2,3,…n = 1, 2, 3, \ldots to get the terms one after another.

Exam­ple 1

Write the first four terms of the sequence with tn=3n−1t_n = 3n - 1.

t1=3(1)−1=2,t2=3(2)−1=5,t3=3(3)−1=8,t4=3(4)−1=11t_1 = 3(1) - 1 = 2, \quad t_2 = 3(2) - 1 = 5, \quad t_3 = 3(3) - 1 = 8, \quad t_4 = 3(4) - 1 = 11

So the sequence is 2,5,8,11,…2, 5, 8, 11, \ldots

Exam­ple 2

Write the first four terms of the sequence with tn=n2+1t_n = n^2 + 1, and find t10t_{10}.

t1=2t_1 = 2, t2=5t_2 = 5, t3=10t_3 = 10, t4=17t_4 = 17. And t10=102+1=101t_{10} = 10^2 + 1 = 101.

Exam­ple 3

Write the first five terms of the sequence with tn=(−1)n⋅nt_n = (-1)^n \cdot n.

When nn is odd, (−1)n=−1(-1)^n = -1; when nn is even, (−1)n=1(-1)^n = 1. So the terms are −1,2,−3,4,−5,…-1, 2, -3, 4, -5, \ldots, with signs that alter­nate.

Describ­ing a sequence by a rule link­ing terms

Some­times each term is built from the terms before it. Such a rule is called a recur­sive rule. For exam­ple:

t1=1,t2=1,tn=tn−1+tn−2 for n≥3t_1 = 1, \quad t_2 = 1, \quad t_n = t_{n-1} + t_{n-2} \text{ for } n \ge 3

Each term is the sum of the two before it, so the sequence is 1,1,2,3,5,8,13,21,…1, 1, 2, 3, 5, 8, 13, 21, \ldots This famous sequence is named after the Ital­ian math­e­mati­cian Fibonacci, and it turns up in the spi­rals of sun­flower seeds and pine cones.

Exam­ple 4

If t1=3t_1 = 3 and tn=2tn−1+1t_n = 2t_{n-1} + 1 for n≥2n \ge 2, write the first four terms.

t1=3t_1 = 3, t2=2(3)+1=7t_2 = 2(3) + 1 = 7, t3=2(7)+1=15t_3 = 2(7) + 1 = 15, t4=2(15)+1=31t_4 = 2(15) + 1 = 31.

Finite and infi­nite sequences

A sequence that stops after a cer­tain num­ber of terms is finite, for exam­ple the two-digit even num­bers 10,12,14,…,9810, 12, 14, \ldots, 98. A sequence that goes on for ever is infi­nite, for exam­ple 1,12,13,14,…\displaystyle 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots The three dots at the end of a list mean that the terms con­tinue in the same way.

Is a given num­ber a term?

To decide whether a num­ber belongs to a sequence, set tnt_n equal to it and solve for nn. The num­ber is a term only if nn comes out as a nat­ural num­ber.

Exam­ple 5

Is 100 a term of the sequence tn=3n+1t_n = 3n + 1? Is 50?

3n+1=1003n + 1 = 100 gives 3n=993n = 99 and n=33n = 33, a nat­ural num­ber. So 100 is the 33rd term.

3n+1=503n + 1 = 50 gives 3n=493n = 49 and n=493\displaystyle n = \frac{49}{3}, which is not a nat­ural num­ber. So 50 is not a term.

Series

The sum of the terms of a sequence is called a series. If t1,t2,t3,…t_1, t_2, t_3, \ldots is a sequence, then the expres­sion

t1+t2+t3+⋯t_1 + t_2 + t_3 + \cdots

is a series. A series is called a finite series if it has a finite num­ber of terms; oth­er­wise it is called an infi­nite series.

The sum of the first nn terms is writ­ten SnS_n:

Sn=t1+t2+⋯+tnS_n = t_1 + t_2 + \cdots + t_n

A short way to write this uses the Greek cap­i­tal let­ter sigma: Sn=∑k=1ntk\displaystyle S_n = \sum_{k=1}^{n} t_k, read "the sum of tkt_k as kk goes from 1 to nn".

Exam­ple 6

For the sequence tn=2n−1t_n = 2n - 1, find S4S_4.

The first four terms are 1,3,5,71, 3, 5, 7. So S4=1+3+5+7=16S_4 = 1 + 3 + 5 + 7 = 16.

From sums back to terms

If you know the sum of the first nn terms for every nn, you can recover each term. The first nn terms add up to SnS_n, and the first n−1n - 1 terms add up to Sn−1S_{n-1}; the dif­fer­ence is the nnth term alone:

tn=Sn−Sn−1(n≥2),t1=S1t_n = S_n - S_{n-1} \quad (n \ge 2), \qquad t_1 = S_1

Exam­ple 7

The sum of the first nn terms of a sequence is Sn=n2+2nS_n = n^2 + 2n. Find the first three terms.

t1=S1=1+2=3t_1 = S_1 = 1 + 2 = 3. Then S2=4+4=8S_2 = 4 + 4 = 8, so t2=S2−S1=8−3=5t_2 = S_2 - S_1 = 8 - 3 = 5. And S3=9+6=15S_3 = 9 + 6 = 15, so t3=15−8=7t_3 = 15 - 8 = 7. The sequence begins 3,5,7,…3, 5, 7, \ldots

Prac­tice

  1. Write the first four terms of the sequence with tn=5n+2t_n = 5n + 2.
  2. Write the first four terms of the sequence with tn=nn+1\displaystyle t_n = \frac{n}{n + 1}.
  3. Find t8t_8 for the sequence with tn=n2−3nt_n = n^2 - 3n.
  4. Write the first five terms of tn=(−1)n+1⋅2nt_n = (-1)^{n+1} \cdot 2n.
  5. If t1=2t_1 = 2 and tn=3tn−1−1t_n = 3t_{n-1} - 1 for n≥2n \ge 2, write the first four terms.
  6. Is 77 a term of the sequence tn=4n−3t_n = 4n - 3? Is 90?
  7. Clas­sify as finite or infi­nite: (a) the mul­ti­ples of 5 between 10 and 100; (b) 1,2,4,8,…1, 2, 4, 8, \ldots
  8. For the sequence tn=3nt_n = 3n, find S5S_5.
  9. If Sn=2n2+nS_n = 2n^2 + n, find t1t_1, t2t_2 and t3t_3.
  10. If Sn=n2+4nS_n = n^2 + 4n, find a for­mula for tnt_n.

Answers

Show answers
  1. 7,12,17,227, 12, 17, 22
  2. 12,23,34,45\displaystyle \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}
  3. t8=64−24=40t_8 = 64 - 24 = 40
  4. 2,−4,6,−8,102, -4, 6, -8, 10
  5. 2,5,14,412, 5, 14, 41
  6. 4n−3=774n - 3 = 77 gives n=20n = 20, so 77 is the 20th term. 4n−3=904n - 3 = 90 gives n=934\displaystyle n = \frac{93}{4}, so 90 is not a term.
  7. (a) finite; (b) infi­nite.
  8. 3+6+9+12+15=453 + 6 + 9 + 12 + 15 = 45
  9. S1=3S_1 = 3, S2=10S_2 = 10, S3=21S_3 = 21, so t1=3t_1 = 3, t2=7t_2 = 7, t3=11t_3 = 11.
  10. tn=(n2+4n)−[(n−1)2+4(n−1)]=2n+3t_n = (n^2 + 4n) - \left[(n - 1)^2 + 4(n - 1)\right] = 2n + 3. This also gives t1=5=S1t_1 = 5 = S_1.