1549901369-Elements_of_Real_Analysis__Denlinger_

8.3 Series with Positive and Negative Terms 483

(c) Let I; an be an absolutely convergent series, and use the same notation

used in Part (a) above. Let c > 0. By the Cauchy criterion for series (8.1.11),

n c

:3 no EN 3 n > m :2:: no==> I: lakl < -.

k=m+l 2

Since the function u : N--+ N is 1-1 and onto, :3 m :2:: n 0 3

{1,2,··· ,no}~ {u1,u2,·· · ,um}·

Suppose n > m. Then,

(12)

(13)

By relation (12) we see that the terms a 1 , a 2 , · · · , an 0 appear in both of the

sums in (13), so they will cancel, leaving only terms with subscript > n 0. Thus,

n:2'.:m==>ISn-Snl=I f ak- ktn aakl'.::'.: I f akl+I ktn aakl

k=no+l ak2no+l k=no+l ak2no+l

n M c c

< I: lakl+ I: iaki<2+2=c,whereM=max{uk:k'.::'.:n}.

k=no+l k=no+l

Therefore, lim Sn - lim Sn = lim (Sn - Sn) = 0. That is,

n-+oo n-+oo n-+oo

00 00

L ak = L aak· •

k=l k=l

With the help of the previous theorem, we can now see one of the deep dif-
ferences between absolute and conditional convergence. The following theorem
shows that this difference is quite dramatic.

Theorem 8.3.14 Given any conditionally convergent series I; an,

(a) I; an has a divergent rearrangement;

(b) Yr E JR, I; an has a rearrangement converging to r.

Proof. Let I: an be a conditionally convergent series. By Theorem 8.3.10
( c), both I; a;i and I; a;;: diverge. After eliminating the zeros from these two
series we can say, in words, that the series of positive terms of I; an diverges
to +oo, and the series of negative terms of I: an diverges to -oo.
(a) Form the desired divergent rearrangement as follows. Since the series
of positive terms diverges, add the first consecutive positive terms until the
sum is greater than 1; then add the first negative term. Then, add the next
consecutive positive terms until the sum is greater than 2, after which add the