1549901369-Elements_of_Real_Analysis__Denlinger_

8.5 Series of Products 495

Thus, \:/ 1 ::; m < n,

n n n

L akbk = L Sk(bk - bk+1) + L (Skbk+I - Sk-1h)

k=m k=m k=m

n

L Sk(bk - bk+I) + ~ - Sm-lbm) + ~ -~

k=m

n
L Sk(bk - bk+I) + Snbn+I - Sm-lbm.
k=m •
The formula in the conclusion of Theorem 8.5.2 is called "summation by
parts" because of its similarity with the integration by parts formula. Admit-
tedly, this similarity is not readily apparent in the form in which it appears
here. To see the simil arity, work Exercise 5.
Abel's summation by parts formula is important because it allows us to
prove Dirichlet's convergence test, which opens the door to the study of trigono-
metric series and other applications.

Theorem 8.5.3 (Dirichlet's Test) Suppose {ak} and {bk} are sequences
such that
n
(a) the sequence {Sn} of partial sums, Sn= I:: ak, is bounded,
k=l
(b) the sequence {bk} is monotone decreasing and nonnegative, and

00

Then I:: akbk converges.

k=l

Proof. Suppose {ak}, {bk}, and {Sn} satisfy the hypotheses. Let c > 0. By

c

(a), :JM> 0 3 't:/n EN, JSnl :SM. By (c), :3 no EN 3 n 2: no:::? 0 :S bn <

2

M.

Then n 0 ::; m < n:::? (by Abel's summation by parts formula)

I

f akbkl =I f Sk(bk - bk+I) + Snbn+l - Smbm+ll

k=m+I k=m+I

n

< L JSkJ (bk - bk+I) + JSnJ bn+I + JSmJ bm+I

k=m+I

= M (bm+l - bn+I + bn+I + bm+I)

= 2Mbm+i < c.