1549901369-Elements_of_Real_Analysis__Denlinger_

64 Chapter 2 11 Sequences

Let n 0 = max{!IJ._, n2l· Then no ?: n 1 and no ?: n2, and so

e e

Ix ~ -L/ < - 2 and /x __?!-O. -Ml < - 2

1 1Xn 0 - LI+ lxn 0 - M/ < e

;.~ I~ -Xn 0 I + lxn 0 - Ml < e

v-·

·1).V~ - IL - Xn 0 + Xn 0 - Ml::=:; IL - Xnol + lxn 0 - M/ < e (Why?)

\"'

/

/L -Xn 0 + Xn 0 - Ml < e

IL-Ml <e.

Thus, Ve> 0, IL - Ml < e. Therefore, by the "forcing principle," L = M.

Theorem 2.2.8 (Alternate Definition of Limit) Xn -> L iff Ve > 0, all
\ ---but finitely many terms of --the sequence - - { -Xn} are in -the interval --(f_. -e, L -+ e).

Proof. (a) (The=? part): Suppose Xn-> L. Let e > 0. By Definition 2.1.4,

3 no E .N 3 n ?: no =? lxn - L/ < e. That is, all terms of the sequence, from the

n 0 th term on, belong to the interval (L - e, L + e). That means that all terms

but the first n 0 - 1 of them are guaranteed to belong to ( L - e, L + e). In other

words, all but finitely many terms of {xn} are in the interval (L -e, L + e).

(b) (The ~ part): Suppose that all but finitely many terms of the sequence

{xn} are in the interval (L-e,L+e). Say all but the first n 1 terms of {xn} are

in the interval (L -e, L + c). That is, n > n 1 =? Xn is in (L - e, L + c). Take

no= n 1 +l. Then, n?: no=? Xn E (L-c, L+c). That is, n?: no=? lxn -LI < e.

Therefore , Xn -> L. •

Definition 2.2.9 A sequence { Xn} is bounded if the set { Xn : n E .N} is a

bounded set. There are two equivalent ways of saying that {xn} is bounded:

(1) 3 a, b E JR 3 \In E .N, a :::; Xn :::; b.

(2) 3 M> 0 3 \In E .N, lxnl:::; M.

Theorem 2.2.10 (Boundedness) Every convergent sequence is bounded.

L+I • 1

------,--------------------------

£ I • • • • • • • •

===~==]==~====~=~==~==~===~=~==~= L-1 • I

Figure 2.2