The Riemann Zeta function, the Riemann Hypothesis and the
distribution of primes.
1. On any compact set K in the region σ > 1 there exists a minimal value of σ, let’s call it σK . Thus
Z ∞
∞ ∞
X
1 X
dx
1
1
≤
< ∞.
≤
1
+
=
1
+
ns σK
σK
n
x
σ
−
1
K
1
n=1
n=1
This implies that the convergence of our Dirichlet series is uniform (the ‘tail’ must go to zero). By theorem
1.1 (p.156) ζ(s) defines an analytic function.
2. Denote the limit superior by σ0 . Let K be a compact subset of the region σ > σ0 . Choose < σK − σ0 .
It follows from the definition of the limit superior that
log |a1 + . . . + an |
< σ0 + log n
for n big enough. In other words: |a1 + . . . + an | < nσ0 + for n big enough. Summation by parts yields
N −1
N
X
an
n=1
a1 + . . . + an X
=
−
(a1 + . . . + an )
ns
Ns
n=1
1
1
−
(n + 1)s ns
We notice that
a1 + . . . + aN ≤ N σ0 +−σK → 0 as N → ∞
Ns
Indeed, we have
Z 1
1
1
1
dx
1
≤ 1
.
(n + 1)s − ns = s
s+1
σK nσK +1
0 (n + x)
(This is just one way of proving the inequality.) Thus
−1
N −1
N
1
1 1 X σ0 +−σK −1
X
(a1 + . . . + an )
−
n
< ∞.
≤
n=1
(n + 1)s ns σK n=1
Again by theorem 1.1 (p.156) the Dirichlet series defines an analytic function for σ > σ0 .
3. Using that only the even terms are left, we get
ζ(s) +
∞
X
(−1)n
n=1
ns
=2
∞
X
n=1
∞
X
1
1 1
2
=2
= s ζ(s).
s
s
s
(2n)
2 n
2
n=1
We must conclude that for σ > 1
∞
X
(−1)n+1
1
.
ζ(s) =
1 − 21−s n=1
ns
On the right hand side is a Dirichlet series with coefficients an = (−1)n+1 . Pluggin this in the limit superior
formula of exercise 2 yields σ0 = 0. Thus the found expression for ζ(s) also makes sense for σ > 0, except
2πi
in s ∈ 1 + log
Z were it clearly has at most simple poles.
2
1
4. We can write the product as
Y
p
Y
1
1
1
=
1 + s + 2 s + ...
1 − p−s
p
(p )
p
Since every natural number has a unique prime decomposition, this must equal
∞
X
1
ns
n=1
5. We take any (possibly multivalued) definition of the complex logarithm. We notice for σ > 1
X
1
1
1
3 log ζ(σ) + 4 log |ζ(σ + it)| + log |ζ(σ + 2it)| = Re
3 log
+ 4 log
+ log
1 − p−σ
1 − p−σ−it
1 − p−σ−2it
p
∞ XX
3p−kσ + 4p−k(σ+it) + p−k(σ+2it)
= Re
k
p k=1
=
∞
XX
p−kσ
p
=2
k=1
∞
XX
p
k=1
k
(3 + 4 cos(kt log k) + cos(2kt log k))
p−kσ
(1 + cos(kt log k))2 ≥ 0.
k
It follows that |ζ(σ)3 ζ(σ + it)ζ(σ + 2it)| ≥ 1 for σ > 1. This proves that ζ(s) has no zeros for σ > 1. Now
suppose suppose ζ(s) has a zero in some point ρ = 1 + iτ , then we can write ζ(s) = (s − ρ)f (s) for some
analytic function f . Since ζ has a simple pole in s = 1 we also have ζ(s) = g(s)/(s − 1) for some analytic
function with g with g(1) 6= 0. But then
ζ(σ)3 ζ(σ + it)4 ζ(σ + 2it) =
(σ − 1 + i(t − τ ))4
f (s)g(σ)ζ(σ + 2it)
(σ − 1)3
When we fix t = τ and take the limit σ ↓ 1 we get a contradiction with |ζ(σ)3 ζ(σ + it)ζ(σ + 2it)| ≥ 1.
6. By the previous exercise −ζ 0 (s)/ζ(s) is a well-defined analytic function for σ ≥ 1 except for s = 1. We
have
ζ 0 (s)
−
=−
ζ(s)
Thus the Ikehara-Wiener theorem gives
P
g 0 (s)
s−1
n≤x
g(s)
(s−1)2
g(s)
s−1
−
=−
g 0 (s)
1
+
.
g(s)
s−1
Λ(n) ∼ x.
7. We notice that
X
p≤x
log p ≤
X
log x = π(x) log x.
p≤x
Combining this with the hint yields
log x
1X
log x
−
(1 − ) π(x)
+ O(x ) ≤
log p ≤ π(x)
x
x p≤x
x
Taking the limit x → ∞ and realising that the inequality is true for any finishes the proof.
2