Step 1. Analytic Properties of the Riemann zeta function
Definition 4.3 A sequence {xn }n≥1 in C is a Cauchy Sequence if, and only
if,
∀ε > 0, ∃N ≥ 1 : ∀n, m ≥ N, |xn − xm | < ε.
Depending on how you have constructed C you may define that a series
converges if, and only if, it is a Cauchy Sequence. Unfortunately this was
not how convergence of a sequence was defined in MATH20101 say.
Theorem 4.4 A sequence {xn }n≥1 converges in C if, and only if, {xn }n≥1
is a Cauchy Sequence.
Proof (⇒) Assume that {xn }n≥1 converges to x say. Let ε > 0 be given.
Then by the ε − N definition of convergence, ∃N ≥ 1 : |xn − x| < ε/2. Let
m, n ≥ N then
|xn − xm | = |xn − x + x − xm | ≤ |xn − x| + |x − xm | < ε,
and we have verified the definition of a Cauchy Sequence.
(⇐) This follows as a defining property of C, basically that C is complete.
Definition 4.5 Let {Fn (z)}n≥1 be a sequence of functions defined on a set
D ⊆ C.
The sequence {Fn }n≥1 converges to F on D iff
∀ε > 0, ∀z ∈ D, ∃N = N (ε, z) : ∀n ≥ N, |Fn (z) − F (z)| < ε.
The sequence {Fn }n≥1 converges uniformly to F on D iff
∀ε > 0, ∃N = N (ε) : ∀z ∈ D, ∀n ≥ N, |Fn (z) − F (z)| < ε.
Equivalently the definition of uniform convergence can be given in terms of
a Cauchy sequence as
∀ε > 0, ∃N = N (ε) : ∀z ∈ D, ∀n, m ≥ N, |Fn (z) − Fm (z)| < ε.
All results of sequences translate as results for series, as in
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Definition 4.6 A series ∞
j=1 fj (z) converges uniformly on D if, and
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only if, the sequence of partial sums Fn (z) = nj=1 fj (z) converges uniformly
on D. That is,
n
X
fj (z) < ε.
∀ε > 0, ∃N = N (ε) : ∀z ∈ D, ∀n, m ≥ N, j=m+1
This leads to
Weierstrass’s M -Test for Uniform Convergence of a Series: If there
exists a sequence {Mi }i≥1 of non-negative numbers for which |fj (z)| < Mj
P∞
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f or all z ∈ D and all i ≥ 1 and ∞
j=1 Mj converges then
j=1 fj (z) converges uniformly on D.
Verification In fact, assuming k > ℓ, the partial sums satisfy
|Fk (z) − Fℓ (z)| = |fk (z) + fk−1 (z) + ... + fℓ+1 (z)|
≤ |fk (z)| + |fk−1 (z)| + ... + |fℓ+1 (z)|
< Mk + Mk−1 + ... + Mℓ+1 .
On account of the convergence of the series the last sum can be made smaller
than a given ε > 0. Hence the sequence of partial sums Fm (z) converges
uniformly.
The Question then arises what properties of a uniformly converging sequence
are inherited by its limit?
We have Weierstrass’s Theorem for Series (see Background Notes 0.7) that
states
Theorem 4.7 Weierstrass’s Theorem for Series. Assume
f1 (z) , f2 (z) , f3 (z) , ...
P∞
are holomorphic in an open set D, and
i=1 fi (z) converge uniformly on
every closed and bounded subset of D. Then
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(i) The F (z) = ∞
i=1 fi (z) is holomorphic on D,
P∞ (k)
(ii) For all k ≥ 1, the series
(z) converges on D, and converge
i=1 fi
uniformly on every closed and bounded subset of D with limit F (k) (z). (So
the series can be differentiated term-by-term.)
6
The major example of use of this theorem is the Riemann zeta function.
The
zeta function is a particular example of a Dirichlet Series,
P∞ Riemann
−s
n=1 an n . For a given s ∈ C such a series may converge or not and if it
converges it may converge absolutely or not. Results on how such regions are
connected for general Dirichlet Series are not
Pgive in this course. We will,
though, make use of the fact that if a series ∞
n=1 an converges absolutely
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P∞
then we have an “infinite” triangle inequality | ∞
a
|
≤
n
n=1
n=1 |an | .
For Re s ≥ 1 + δ the Riemann zeta function
absolutely and to
P∞ converges
s
see this use the Comparison Test on
n=1 |1/n |, for each term satisfies
P∞
s
1+δ
1+δ
|1/n | ≤ 1/n
and n=1 1/n
converges.
To go further we need to show that the series defining ζ (s) converges uniformly on some domain.
Theorem 4.8 Assume δ > 0. For Re s ≥ 1 + δ the Riemann zeta function
∞
X
1
ζ (s) =
ns
n=1
converges uniformly, is holomorphic in Rs > 1, with derivative
′
ζ (s) = −
∞
X
log n
n=1
ns
for Re s > 1.
Proof We have to show only that the series defining ζ (s) converges uniformly
Re s ≥ 1 + δ. We could simply apply Weierstrass’s M -test with Mn = 1/n1+δ
or do it directly as
N
∞
∞
X
X
X
1
1 1 =
≤
ζ (s) −
ns s
s
n
n
n=1
n=N +1
n=N +1
by “infinite” triangle inequality
≤
∞
X
1
n=N +1
≤
Z
∞
N
n1+δ
since Re s ≥ 1 + δ
du
1
=
.
1+δ
u
δN δ
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Given any ε > 0, then 1/δN δ is less than ε when N is sufficiently large,
independent of s, showing uniform convergence. We can apply Weierstrass’s
Theorem since any closed and bounded subset of Re s > 1 is contained in
Re s ≥ 1 + δ for some δ > 0. Each term in the series is holomorphic on C
with derivative
d
1
d
log n
= e−s log n = − s .
s
ds n
ds
n
The stated result then follows from Weierstrass’s Theorem.
Earlier in the course we looked at replacing sums by integrals. In the same
vein we can replace the Dirichlet Series defining the Riemann zeta-function
by an integral.
Theorem 4.9 For Re s > 1,
1
−s
ζ (s) = 1 +
s−1
Z
∞
1
{u}
du
u1+s
(4)
where {u} is the fractional part of u (so {u} = u − [u]). Further, the right
hand side of (4) defines a function holomorphic in Re s > 0, except for a
simple pole, residue 1 at s = 1.
This is an example of an analytic continuation of ζ (s), a function holomorphic on a region containing Re s > 1, on which it agrees with the Dirichlet
series definition of ζ (s). It is in fact possible to continue analytically ζ (s)
to all of C, though we do not do so in this course.
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Proof Recall the notation [x] = 1≤n≤x 1 so {x} = x − [x]. By Partial Summation (and this argument, for s = 1, has been seen before in the previous
Chapter)
X 1
X 1
1
1
=
−
− s
s
s
s
n
N
N
n
1≤n≤N
1≤n≤N
X Z N
du
N
(−s) s+1
−
=
s
N
u
1≤n≤N n
N
+s
=
Ns
Z
N
=
+s
Ns
Z
N
+s
=
Ns
Z
N
1
X
1≤n≤u
N
[u]
1
N
1
1
du
us+1
du
us+1
du
u s+1 − s
u
Z
N
(u − [u])
1
N
s
=
+
N 1−s − 1 − s
s
N
1−s
Z
N
{u}
1
du
us+1
du
.
us+1
(5)
Remember that Re s = σ > 1. Let N → ∞, then
1−s N = N 1−σ → 0
while
{u} 1
u1+s ≤ u1+σ
and so the integral in (5) converges. Hence
Z ∞
1
{u}
ζ (s) = 1 +
−s
du
s−1
u1+s
1
(6)
(7)
as required. In fact, from (6), we see that the integral in (7) converges
absolutely for Re s > 0.
There is a theory for uniform convergence of integrals; see Weierstrass’s Theorem for Infinite Integrals in Background Notes 0.7. But this is not directly
applicable here since the integrand {u} u−1−s is not a continuous function of
u for fixed s.
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Instead, consider the infinite integral in (7) as an infinite sum of integrals
over intervals of length 1:
Z
∞
1
∞
X
{u}
du
=
u1+s
n=1
Z
n+1
n
∞
{u} du X
=
fn (s) .
u1+s
n=1
We can show (but will only do so for level 4) that for any δ > 0,
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(i) the series ∞
n=1 fn (s) converges uniformly on Re s > δ,
(ii) each fn (s) is holomorphic on Re s > δ, with derivative
Z n+1
{u}
dfn (s)
=
(− log u) du.
ds
u1+s
n
This is sufficient, by Weierstrass’s Theorem for Series, part i of Theorem 4.7
above, to imply that the integral in (7) is holomorphic on Re s > 0.
We can further use part ii of Theorem 4.7 to justify the differentiating seen
below of the series term-by-term,
d
ds
Z
∞
1
∞
{u}
d X
du
=
fn (s)
u1+s
ds n−1
∞
X
d
fn (s)
=
ds
n−1
= −
∞ Z
X
n=1
= −
Z
∞
1
n+1
n
{u}
log udu
u1+s
{u}
log udu.
u1+s
Immediate from the proof, and in particular (14), is
Corollary 4.10 For Re s > 0,
1
ζ (s) = −
−
(s − 1)2
′
Z
∞
1
{u}
du + s
us+1
10
Z
∞
1
{u} log u
du.
us+1