## Infinite Sequences

Consider the function / defined for n = 1, 2, 3,... by the formula f(n) = l/n. Then /(1) = 1, /(2) = 1/2, /(3) = 1/3, and so on. The list of numbers

1111 M

is called an infinite sequence. Its first term is 1, and its general (nth) term is \/n. In general, any function whose domain is the entire set of positive integers is called an infinite sequence. Similarly s„ = 100 • 1.08n~x(n = 1,2,...) determines an infinite sequence whose first terms are

100, 100-1.08, 100 • 1.082, 100 • 1.083, ... [**]

If 5 is an infinite sequence, its terms s(l), 5(2). 5(3),____5(n),... are usually denoted by using subscripts: s\, s2, 53, ..., sn,____We use the notation {5„}^i,, or simply {5„}, for an arbitrary infinite sequence.

Consider the previous sequence [*]. If we choose n large enough, the terms can be made as small as we like. We say that the sequence converges to 0. In general, we introduce the following definition:

A sequence {s„} is said to converge to a number s if s„ is arbitrarily close to s for all n sufficiently large. We write

A sequence that does not converge to any real number is said to diverge. For example, the sequence in [**] earlier diverges because 100 ■ 1.08"-1 tends to oo as n tends to oo.

The definition of convergence for a sequence is a special case of the previous definition that f(x) —► A as x oo. All the ordinary limit rules in Section 4.4 apply to limits of sequences.

Example 6.10

Write down the first five terms of the following sequences:

jV + l| 2 9 28 65 126 (C) \n2 + 2) 1 3' 6' IT 18' 27 '

The sequence in (a) converges to 0, because 1 jn tends to 0 as n tends to oo. The sequence in (b) converges to 3, because (1/10)" tends to 0 as n tends to oo. The sequence in (c) is divergent. To see this note that lim sn — s or sn s as n —> oo

Then decide whether or not each converges. Solution

Example 6.11

For n > 3 let An be the area of a regular «-polygon inscribed in a circle with radius 1. For n = 3, A3 is the area of a triangle; for n = 4, A4 is the area of a square; for n = 5, A5 is the area of a pentagon; and so on (see Fig. 6.14).

The larger n is, the larger is An, but each An is less than 7r, the area of a circle with radius 1. It seems intuitively evident that we can make the difference between An and rz as small we wish if only n becomes sufficiently large, so that

In this example, A\ and A2 have no meaning, so the sequence starts with A3.

The sequence {A„} in the previous example converges to the irrational number it = 3.14159265----Another sequence that converges to it starts this way: s\ —

3.1, 52 = 3.14, S3 = 3.141, 54 = 3.1415, etc. Each new number is obtained by including an additional digit in the decimal expansion for n. For this sequence, sn —> 71 as /2 —> 00.

Consider an arbitrary irrational number r. Just as for tt, the decimal expansion of r will define one particular sequence rn of rational numbers that converges to r. Actually, each irrational number is the limit of infinitely many different sequences of rational numbers.

### Example 6.12

It is often difficult to determine whether or not a sequence is convergent. For example, consider the sequence whose general term is sn = (1 4- 1 /rif. Do you think that this converges? The values of sn for some values of n are given by n 1 2 3 5 10 100 10,000 100,000

(1 + Tj)" 2 2.25 2.37 2.49 2.59 2.70 2.7181 2.7182

This table seems to suggest that s„ tends to a number close to 2.718. One

FIGURE 6.14

as n —> co can prove that {s„} does converge by relying on the general property. Any increasing sequence of real numbers that has an upper bound is convergent. The limit of {s„} is an irrational number denoted by e, which is one of the most important constants in mathematics. See Section 8.1.

Problems

Find the following limits:

n-*-oo n-*oo n-*oo d. lim a„ßn e. lim a„/ß„ f. lim yjßn-ctn n—*oo n—+ 00 n~oo

2. Examine the convergence of the sequences whose general terms are as follows: