Advanced High-School Mathematics

SECTION 5.2 Numerical Series 281

∑n

k=1

akbk = snbn+1+

∑n

k=1

sk(bk−bk+1).

Proof. Settings 0 = 0 we obviously haveak=sn−sk− 1 , k≥1. From
this, one has

∑n

k=1

akbk =

∑n

k=1

(sk−sk− 1 )bk

=

∑n

k=1

skbk−

∑n

k=1

sk− 1 bk

= snbn+1+

∑n

k=1

sk(bk−bk+1).

Dirichlet’s Theorem for Convergence. Let (an) and (bn) be

two sequences of real numbers. If the partial sums

∣∣

∣∣

∣

∑n

k=1

an

∣∣

∣∣

∣ are all

bounded by some constantM, and if

b 1 > b 2 > b 3 >···≥ 0 with nlim→∞bn= 0,

then the series

∑∞

k=1

akbk converges.

Proof. Settingsn=

∑n

k=1

anandrn=

∑n

k=1

akbk we have from the above

lemma that

rn−rm=snbn+1−smbm+1+

∑n

k=m+1

sk(bk−bk+1),

wheren≥mare positive indices. Taking absolute values and applying
theTriangle inequalityprovides

|rn−rm| ≤ |sn|bn+1+|sm|bm+1+

∑n

k=m+1

|sk|(bk−bk+1)

≤ Mbn+1+Mbm+1+M

∑n

k=m+1

(bk−bk+1)

= 2Mbm+1.