Theorem 1 (Euler)

Theorem 5.

lim inf
x→∞


π(x) x/ log x



≤ ≤
1 lim sup
x→∞


π(x)
.
x/ log x

n/ log n
Theorem 5 implies that, if the limit lim_n→∞ ^π(n) exists, then it is equal to 1.
We will prove instead that

lim inf
x→∞
ψ(x) x



≤ ≤
1 lim sup
x→∞
ψ(x)
,
x

ζ(s)
which by Theorem 2 is equivalent to the assertion in Theorem 5. Let f (s) = − ^ζ′^(s) for every real s > 1, and let

l = lim inf
x→∞
ψ(x) x


, L = lim sup
x→∞
ψ(x) x

l^′ = lim inf(s − 1)f (s) , L^′ = lim sup(s − 1)f (s)

Obviously l ≤ L and l^′ ≤ L^′.
s→1⁺
s→1⁺

We will show that l ≤ l^′ ≤ L^′ ≤ L and l^′ = L^′ = 1; together these two give us Theorem 5.

x
Choose some arbitrary real B > L, then ^ψ(x) < B for all x ≥ x₀ = x₀(B), and we can assume without loss
of generality x₀ > 1.
Here we will need a statement of Abel’s Theorem, whose proof we will omit:



Σ
Theorem 6. Let 0 ≤ λ₁ ≤ λ₂ ≤ . . . be a sequence of real numbers, such that λ_n → ∞ as n → ∞ and let (a_n) be a sequence of complex numbers. Let A(x) = _λn≤x a_n, and ϕ(x) a complex-valued function defined for x ≥ 0. Then
k k−1
^Σ a_nϕ(λ_n) = A(λ_k)ϕ(λ_k) − ^Σ A(λ_n)(ϕ(λ_n+1) − ϕ(λ_n)). (22)



n=1

n=1

∫
If ϕ has a continuous derivative in (0, ∞), and x ≥ λ₁, then equation 22 can be written as

^λΣn^≤x
a_nϕ(λ_n) = A(x)ϕ(x) −


x
A(t)ϕ^′(t)dt. (23)
^λ1

If, in addition, A(x)ϕ(x) → 0 as x → ∞, then

∞
nΣ=1


a_nϕ(λ_n
) = ∞ A(t)ϕ^′(t)dt, (24)

∫−
^λ1

provided either side is convergent.

Now for s > 1, set λ_n = n, a_n = Λ(n), and ϕ(x) = x^−s, then A(x) = ψ(x), and A(x)ϕ(x) → ∞ as x → ∞, since ψ(x) ≤ π(x) log x ≤ x log x, so that A(x)ϕ(x) = O(x^1−s log x) = o(1). We will also use without proof that

Σ₋
ζ^′(s) ₌ ^∞
ζ(s)
n=1
Λ(n)
_ns , (s real, s > 1).

Combining this with equation 24 gives
^{∫ ∞} ψ(x)

_x

^{∫ x0} ψ(x)

_x

∫ _{∞ B}

so that
f (s) = s
1
_s+1 dx < s
1
_s+1 dx +
^x0
_xs dx,

^{∫ x0} ψ(x)

_x

∫ _{∞ B}

_x

^{∫ x0} ψ(x)
sB

∫
We set
f (s) < s
1
_s+1 dx + s
1
_s dx < s
1
_x2 dx + _{s − 1} .

^x0 ψ(x)
_x2 dx = K = K(x₀) = K(x₀, B),
1
so that we can rewrite the above equation in the form

We let s → 1⁺, obtaining
(s − 1)f (s) < s(s − 1)K + sB.

— ≤
L^′ = lim sup(s 1)f (s) B,
s→1+0

and since B > L was arbitrary we have necessarily L^′ ≤ L. An analogous argument gives l ≤ l^′, and so
l ≤ l^′ ≤ L^′ ≤ L
Now we prove l^′ = L^′ = 1 by showing

lim
s→1⁺
(s 1)²ζ^′(s) = 1 and lim (s 1)ζ(s) = 1.

— − −
s→1⁺

This implies (s − 1)f (s) → 1 as s → 1⁺.
For s > 1, we have that x^−s is decreasing as a function of x, and so

^{∫ ∞} dx _< ^Σ∞ 1
< 1 + ∫ ∞ ^dx ;

hence
_{1 x}s

_ns
n=1
_{1 x}s

implying (s − 1)ζ(s) → 1 as s → 1⁺.

1


s − 1


< ζ(s) <
s
,
s − 1

Now for the second part we have that, for s > 1 and x ≥ e, the function x^−s log x is decreasing, so

∞
−ζ^′(s) = ^{Σ log n} = ∫ ∞ ^{log x}dx + O(1).

By substituting x^s−1 = e^y we get

_ns
n=1
_{1 x}s

−ζ^′(s) = ¹ ∫ ∞ ye^−ydy + O(1) = ¹ + O(1),

so that

(s − 1)²
0

(s − 1)²ζ^′(s)

(s − 1)²

+

(s − 1)f (s) = − _{(s − 1)ζ(s)} → 1 as s → 1 .
.
This means that l^′ = L^′ = 1 and thus l ≤ 1 ≤ L, concluding the proof of the Theorem.