The Real And Complex Number Systems

(a) If fn is multiplicative and if the series ∑ fn converges absolutely, prove that
∑
n1
fn 1 fp k fp k2 ...,
k1
where p k denote the kth prime, the product being absolutely convergent.
Proof: We consider the partial product P m m
k1
1 fp k fp k2 ... and show
that P m → ∑
n1
fn as m → . Writing each factor as a geometric series we have
m
P m 1 fp k fp k2 ...,
k1
a product of a finite number of absolutely convergent series. When we multiple these series
together and rearrange the terms such that a typical term of the new absolutely convergent
series is
fn fp 1
a 1
fp m
a m
,wheren p 1
a 1
p m
a m
,
and each a i ≥ 0. Therefore, we have
P m ∑ fn,
1
where ∑ 1
is summed over those n having all their prime factors ≤ p m .Bytheunique
factorization theorem (Theorem 1.9), each such n occors once and only once in ∑ 1
.
Substracting P m from ∑
n1
fn, weget
∑ fn − P m ∑ fn − ∑ fn ∑ fn
n1
n1
1
2
where ∑ 2
is summed over those n having at least one prime factor p m . Since these n
occors among the integers p m ,wehave
∑
n1
fn − P m
≤ ∑
np m
|fn|.
As m → the last sum tends to 0 because ∑ n1
fn converges, so P m → ∑ n1
fn.
To prove that the product converges absolutely we use Theorem 8.52. The product has
the form 1 a k ,where
a k fp k fp k2 ....
The series ∑|a k | converges since it is dominated by ∑ n1|fn|.
Thereofore, 1 ak
also converges absolutely.
Remark: The method comes from Euler. By the same method, it also shows that there
are infinitely many primes. The reader can see the book, An Introduction To The Theory
Of Numbers by Loo-Keng Hua, pp 91-93. (Chinese Version)
(b) If, in addition, fn is completely multiplicative, prove that the formula in (a)
becomes
∑ fn
1
1 − fp
n1
k1
k .
Note that Euler’s product for s (Theorem 8.56) is the special case in which fn n −s .
Proof: By(a),iffn is completely multiplicative, then rewrite