1549901369-Elements_of_Real_Analysis__Denlinger_

2.3 Inequalities and Limits 73

Outline of a proof of (a) [fill in missing steps, and justify]:

c:.

Case 1 (L = 0): Then 3n 1 EN 3 n ~ n1 =? lxnl <

2

. Holdmg n1 fixed,
X1 + X2 + · · · + Xn lx1 + X2 + · · · + Xn I
---------=-^1 --> 0, so :3n2 EN 3 n ~ n 2 =?^1 <
n n
c:
2· For n ~no= max{n1, n2}, lcrnl =

lx1+x2+···+xn 1 I lxn 1 +1+Xn 1 +2+···+xnl C: 1 C:

n + n <2+;(n-ni)2

< c:.

Case 2 (L "I-0): Apply Case 1 to the sequence {xn - L}.

25. (Project) Geometric Means: For a given sequence { xn} we define its
    sequence of geometric means to be { Tn}, where Tn = y'x 1 x2 · · · Xn·

(a) Prove that if Xn--> L, then Tn--> L.

(b) Prove that the converse of (a) is false, by finding a sequence {xn}

such that {Tn} converges but {xn} does not.

2.3 Inequalities and Limits

The following theorem and its corollary provide two very useful tools for proving
that sequences converge to a limit L. You will find yourself using these tools,
especially the second, quite often.

Theorem 2.3.1 (The First "Squeeze" Principle) If {an}, {bn}, and {en}
are sequences such that an --> L, Cn --> L, and Vn E N, an :::; bn :::; Cn, then
bn--> L.

-L-

a" bll

Figure 2.4

Proof. Suppose {an}, {bn}, and {en} are convergent sequences such that
an --> L, Cn --> L, and Vn E N, an :::; bn :::; Cn. Let c: > 0.
Since an--> L , 3 n1 EN 3 n ~ n1 =? Ian - LI < c:.
Since Cn--> L, 3 n2 EN 3 n ~ n2 =? len - LI < c:.