P
1
p≤n p
= ln(ln(n) + O(1): An Exposition
by William Gasarch and Larry Washington
1
Introduction
P
It is well known that p≤n p1 = ln(ln(n)) + O(1) where p goes over the primes. We
give several known proofs of this.P
We first present a proof that p≤n p1 ≥ ln(ln(n)) + O(1). This is based on Euler’s
P
P
proof that p p1 diverges. We then present three proofs that p≤n p1 ≤ ln(ln(n)) +
O(1). The first one, essentially due to Mertens, does not use the prime number
theorem. The second and third one do use the prime number theorem and hence are
shorter.
P
For a complete treatment of Merten’s proof that p p1 diverges, and how it compares with modern treatments, see the scholarly work of Villarino [4].
2
Euler’s Proof that
1
p≤n p
P
≥ ln(ln(n)) + O(1)
The proof here follows the one in [1].
Lemma 2.1 For 0 ≤ x ≤ 1/2, − ln(1 − x) ≤ x + x2 .
Rx 1
1
dt. For 0 ≤ t ≤ 1/2, 1−t
≤ 1 + 2t. Hence
Proof:
− ln(1 − x) = 0 1−t
Z x
Z x
1
(1 + 2t)dt = x + x2 .
− ln(1 − x) =
dt ≤
0 1−t
0
Theorem 2.2
Proof:
≥ ln(ln(n)) + O(1).
Clearly
∞
X
1
j=1
1
p≤n p
P
j
= (1 −
1
1
1
1
1
1
+ 2 + · · · )(1 − + 2 + · · · ) · · · =
×
× ···
2 2
3 3
1 − 2−1 1 − 3−1
which we rewrite as
∞
X
1
j=1
j
=
Y
(1 − p−1 )−1
p
We need a finite version of this statement. Let Sn be the set of natural numbers
whose prime factors p are all ≤ n. Then
X1 Y
=
(1 − p−1 )−1 .
j
p≤n
j∈S
n
1
Clearly
1
j≤n j
P
≤
1
j∈Sn j .
P
By integration ln n ≤
ln(n) ≤
X1
j≤n
≤
j
1
j≤n j .
P
Hence we have
X1 Y
=
(1 − p−1 )−1
j p≤n
j∈S
n
ln(ln(n)) ≤
X
− ln(1 − p−1 ).
p≤n
By Lemma 2.1
X
− ln(1 − p−1 ) ≤
p≤n
X1
p≤n
p
+
1
.
p2
Putting this all together we get
X1
p≤n
p
≥ ln(ln(n)) −
Since the second sum is bounded by
X1
p≤n
p
P∞
1
i=1 i2 ,
X 1
p2
p≤n
which converges, we have
≥ ln(ln(n)) − O(1).
Note 2.3 If the abovePproof is done more carefully with attention paid to the constants you can obtain p≤n p1 ≥ ln(ln(n)) − 0.48. See [1].
3
Mertens Proof that Does Not Use the Prime
Number Theorem
This is adapted from Landau’s book [2]. He works a little harder and gets o(1) instead
of O(1).
We first need a weak form of the prime number theorem.
Lemma 3.1 π(x) = O(x/ ln x).
Proof:
Let n be a positive integer. Clearly every prime
p with n < p ≤ 2n occurs
. Therefore,
in the prime factorization of the binomial coefficient 2n
n
Y
Y
2n
π(2n)−π(n)
n
=
n≤
p≤
≤ (1 + 1)2n = 4n .
n
n<p≤2n
n<p≤2n
2
Taking logs yields
n ln 4
≤ 2 ln 4
π(2n) − π(n) ≤
ln n
2n
n
−
ln(2n) ln n
for n ≥ 8. If y ≥ 16 is a real number, let 2n be the largest even integer with 2n ≤ y.
Then π(y) − π(2n) ≤ 1 and |π(y/2) − π(n)| ≤ 1. By increasing 2 ln 4 to 4 we can
absorb these errors and obtain
y/2
y
−
π(y) − π(y/2) ≤ 4
ln y ln(y/2)
for y ≥ 16. Adding up this inequality for y = x, x/2, x/4, . . . yields
x π(x) − π(16) ≤ 4
.
ln x
This yields the lemma.
We now need a result that is interesting in its own right.
Proposition 3.2
X ln p
p≤x
Proof:
p
= ln x + O(1).
If n is a positive integer and p is a prime, the power of p dividing n! is
n
n
n
+ 2 + 3 + ··· .
p
p
p
Therefore,
n
n
n
ln(n!) =
ln p
+ 2 + 3 + ··· .
p
p
p
p≤n
X
Changing bn/pc to n/p introduces an error of most 1, so we have
X
X ln p
X
n
=n
+ O(
ln p).
ln p
p
p
p≤n
p≤n
p≤n
Since there are π(n) terms in the sum, Lemma 1 implies that
X
O(
ln p) = O(π(n) ln n) = O(n).
p≤n
Let’s treat the higher terms:
n
n
n
n
−1
−2
+
+
·
·
·
<
1
+
p
+
p
+
·
·
·
= 2
.
2
3
2
p
p
p
p −p
3
Therefore,
X ln p
n
n
+
+
·
·
·
≤
n
= O(n)
ln p
2
3
2−p
p
p
p
p≤n
p≤n
P
P
2
2
since
ln p/(p − p) ≤ j≥2 ln j/(j − j), which converges.
Stirling’s formula says that
X
ln(n!) = n ln n + O(n)
R
P
(this weak form can be proved by comparing ln j with ln t dt). Putting everything
together yields
X ln p
n ln n + O(n) = n
+ O(n).
p
p≤n
Dividing by n yields the proposition for x = n. The error introduced by changing
from x to n = bxc is absorbed by O(x), so the proposition is proved.
The following lemma is well known. It is an analog of integration by parts for
summations. It is easily proven by induction on n.
Lemma 3.3 Let both f1 , f2 , . . . and g1 , g2 , . . . be sequences of complex numbers. Then,
for all m ≤ n,
n
X
fi (gi+1 − gi ) = fn+1 gn+1 − fm gm −
i=m
n
X
gi+1 (fi+1 − fi ).
i=m
We can now prove the theorem.
Theorem 3.4
Proof:
1
p≤x p
P
= ln ln x + O(1).
We have
f (x) =
X ln p
p≤x
p
= ln x + r(x),
where r(x) = O(1). Then
x
X ln p 1
X
f (n) − f (n − 1)
=
=
p p≤x p ln p n=2
ln n
X1
p≤x
=
x
X
ln n − ln(n − 1)
n=2
Since
ln n
+
x
X
r(n) − r(n − 1)
n=2
log n
1
1
ln n − ln(n − 1) = − ln 1 −
= + O(1/n2 ),
n
n
4
.
and
x
X
n=2
we find that
1
= ln ln x + O(1),
n ln n
x
X
ln n − ln(n − 1)
n=2
ln n
= ln ln x + O(1).
Summation by parts yields
x
X
r(n) − r(n − 1)
log n
n=2
x
X
1
1
r(bxc)
=
r(n)
−
+
ln n ln(n + 1)
ln(bxc + 1)
n=2
!
x
X
1/n
=O
+ O(1) = O(1).
2
(ln
n)
n=2
Putting everything together yields the theorem.
4
A Proof that uses Summation by Parts
In this section we give the standard way to estimate
Theorem.
P
1
Theorem 4.1
p≤n p = ln(ln(n)) + O(1).
P
1/p using the Prime Number
Proof:
Let π(i) be the number of primes ≤ i. Let g(i) = π(i − 1) and f (i) = 1i .
Let m = 2. Plugging these into Lemma 3.3 yields
n
X
1
i=2
n
X
1
1
1
1
(π(i) − π(i − 1)) =
π(n) − π(1) −
π(i)(
− ).
i
n+1
2
i+1
i
i=2
We need:
• π(i) − π(i − 1) is 1 if i is prime but 0 otherwise.
• π(n) = lnnn + O( lnn2 n ) by the Prime Number Theorem (when it is proved with
an error term).
We have
1
1
π(i)
1
π(i)(
− )=
=
+O
i+1
i
i(i + 1)
(i + 1) ln i
1
(i + 1) ln2 i
by the Prime Number Theorem. But this equals
1
1
1
1
1
=
.
−
+O
+O
i ln i i(i + 1) ln i
i ln i
(i + 1) ln2 i
(i + 1) ln2 i
5
Therefore,
n
X
1
1
= ln(ln(n)) + O(1),
=
+O
p
i ln i
(i + 1) ln2 i
i=2
X1
p≤n
where we have used
Z n
n
X
1
1
=
dx + O(1) = ln(ln(x)) + O(1)
i
ln
i
x
ln
x
2
i=2
and
n
X
i=2
1
= O(1)
(i + 1) ln2 i
by the Integral Test.
5
A Proof that uses Integration by Parts
This is the same as the previous proof, with the summation by parts replaced by
integration by parts in a Stieltjes integral.
Theorem 5.1
Proof:
1
p≤n p
P
= ln(ln(n)) + O(1).
The preceding proof can be rewritten using Stieltjes integrals:
X1 Z x 1
=
dπ(t).
p
t
1.9
p≤x
Integration by parts yields
π(x)
+
x
Z
x
1.9
π(t)
dt.
t2
We use the Prime Number Theorem approximation π(x) = lnxx + O( lnx2 x ) to obtain
Z x
Z x
1
1
1
+
+ O(
2 ) = ln(ln(x)) + O(1).
ln x
1.9 t ln t
1.9 t ln t
6
6
What Else is Known
Rosser and Schoenfeld [3] have shown that, when n ≥ 286,
ln(ln n) −
X1
1
1
≤
ln(ln
n)
+
+
B
≤
+ B,
2
2(ln n)2
p
(2
ln
n)
p≤n
where B = 0.261497212847643.
P
Even though the sum p≤n
P
1
•
p≤10 p = 1.176
P
1
•
p≤106 p = 2.887
P
1
•
p≤109 p = 3.293
P
1
•
p≤10100 p ∼ 5.7
1
p
diverges, it grows very slowly:
References
[1] J. Kraft and L. Washington. An introduction to Number Ttheory and Cryptogaphy.
CRC Press, 2014.
[2] E. Landau. Handbuch der Lehre von der Werteilung der Primzahlen. Chelsea
Publhsing Co, 1953.
[3] J. Roser and L. Schoenfeld. Approximate formulas for some properties of prime
numbers. Illinois Journal of Mathematics, pages 64–94, 1962.
[4] M. B. Villarino. Merten’s proof of Mertens’ theorem, 2005. http://arxiv.org/
abs/1205.3813.
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