1549901369-Elements_of_Real_Analysis__Denlinger_

490 Chapter 8 • Infinite Series of Real Numbers

Note that
(a) Each ck is obtained by grouping k successive terms of the series I: dk.
(b) Each Cn is a partial sum of the series I: dk, and is the sum of terms
occupying a triangle in the upper left corner of the matrix (14) above.
(c) Each AnBn is a sum (but not, strictly speaking, a "partial sum" ) of
terms of I: dk, with terms occupying a square in the upper left corner of the
matrix (14) above.
(d) Every Dn is the sum of terms contained within some square in the
upper left corner of (14) above, and also contained within some triangle in the
upper left corner of (14).
Case 1: Suppose I: ak and I: bk are nonnegative series. By note (d)
above, every Dn is less than or equal to some AmBm. Since I: ak and I: bk are
nonnegative series, we have for all n,

Dn :'.S m--+oo lim (AmBm) = ( m-+oo lim Am) ( m-+oo lim Bm)

i.e., Dn :S (2.::: ak) (2.::: bk)·

Since limits preserve inequalities,

I: dk :::; (I: ak) (2.::: bk)·

On the other hand, by note (c) above, we have for all n,

AnBn :'.SL dk, SO

( n-+oo lim An) ( n-+oo lim Bn) :S L dk

Therefore, putting (15) and (16) together, we have

By note (b) above, {Cn} is a subsequence of {Dn}, so

n-+oo lim Cn = n-+oo lim Dn

(15)

(16)

Case 2: Suppose I: ak and I: bk are not both nonnegative series. By
applying the argument of Case 1 to the series I: lakl, I: lbkl, and I: ldkl, we
conclude that L dk converges absolutely.
Now, bynote(b),thesequence{Cn}isasubsequenceof{Dn},so lim Cn =
n-+oo

n-+oo lim Dn. That is, I: Ck exists and equals I: dk. Also, by note ( c), each AnBn