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.

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

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:

Was this article helpful?

## Post a comment