# AN INEQUALITY INVOLVING PRIME NUMBERS by ddh19362

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```									Univ. Beograd. Publ. Elektrotehn. Fak.
Ser. Mat. 11 (2000), 33–35.

AN INEQUALITY
INVOLVING PRIME NUMBERS
¸
Laurentiu Panaitopol

From Euclid’s proof of the existence inﬁnitely many prime numbers one can
deduce the inequality
p1 p2 · · · pn > pn+1 ,
where pk is the k-th prime number.
Using elementary methods, Bonse proves in [1] that
2
p1 p2 · · · pn > pn+1 for n ≥ 4,

and
3
p1 p2 · · · pn > pn+1 for n ≥ 5.

´
Stronger results of the same nature have been obtained by J. Sandor in [2].
For example
2        2
p1 p2 · · · pn > pn+5 + p [n/2] for n ≥ 24.

Without the restrictions imposed by the use of elementary methods the pre-
´
cise determination of the margin from which the inequality holds, L. Posa [3]
proves the following result:
For all k > 1 there is an nk such that
k
p1 p2 · · · pn > pn+1 for all n ≥ nk .

The aim of the present note is to improve this inequality. We recall two
results due to Rosser and Schoenfeld [5]:

1
(1)                    pn ≤ n log n + log log n −            for n ≥ 20
2

1991 Mathematics Subject Classiﬁcation: 11A41

33
34                                        ¸
Laurentiu Panaitopol

and
x       x
(2)                    π(x) >            +         for n ≥ 59,
log x 2 log2 x
where we denoted by π(x) the number of prime numbers not exceeding x. We will
also use the following result due to G. Robin [4]:
log log n − a
(3)        θ(pn ) > n log n + log log n − 1 +                             for n ≥ 3,
log n
where a = 2.1454 and θ(x) =           log p, the sum being taken after primes p.
p≤x
All these results allow as to prove the following
Theorem. For n ≥ 2
n−π(n)
p1 p2 · · · pn > pn+1       .
We begin by proving the following
Lemma. For n ≥ 59 we have
log log n − 0.4
log pn+1 < log n + log log n +                        .
log n
Proof. It is well known that log x ≤ x − 1 for x > 0, from which we get for
x = 1 + 1/n that
1
log (n + 1) < log n +
n
and then
1                                  1                             1
log log (n + 1) < log log n +        = log log n + log    1+                  < log log n +           .
n                               n log n                       n log n

We apply the inequality (1) and for n ≥ 19 we get
1
log pn+1 < log (n + 1) + log log (n + 1) + log log (n + 1) −
2
1                         1 + log log n       1         1
< log n +  + log log n + log 1 +                 +       2  −
n                            log n        n log n    2 log n
1               log log n      1          1          1
< log n + + log log n +            +          +           −        .
n                 log n     n log n n log2 n 2 log n
It remains to show that
log n + 1                  1    1
+ log log n +        − < log log n − 0.4,
n                   n log n 2
that is
log n + 1      1
+         < 0.1,
n       n log n
which holds for n ≥ 59.
An inequality involving prime numbers                                  35

Proof of the theorem. For n ≥ 59 we use (2) and the Lemma. We have
1       1                                  log log n − 0.4
n−π(n) log pn+1 < n 1 −           −                 log n + log log n +                      .
log n 2 log2 n                                    log n
In order to prove the theorem it is enough to show, using (3), that
1   1                   log y − 0.4                         log y − a
1−     − 2       y + log y                 < y + log y − 1 +               ,
y 2y                         y                                  y
where y = log n > log 59. This last inequality is equivalent to
1               log y − 0.4
a − 0.9 <    1+           log y +                 ,
2y                    y
which is true, since a − 0.9 < 1.3 and log y > log log 59 > 1.4. The theorem is thus
proved for n ≥ 59. It may be checked directly that the assertion in the statement
also holds for 2 ≤ n ≤ 58.
From the above result we immediately obtain an improvement of L. Posa’s  ´
inequality.
Corollary. For any integer k, k ≥ 1, and n ≥ 2k the following inequality holds:
k
p1 p2 · · · pn > pn+1 .

Proof. The function f : N∗ → N, f (n) = n − π(n) in increasing. For n ≥ 2k,
f (n) ≥ f (2k) = 2k − π(2k) ≥ k, since π(2k) ≤ k for k ∈ N∗ .

REFERENCES

1. H. Rademacher, O. Toeplitz: The enjoyment of mathematics. Princeton Univ.
Press, 1957.
´    ¨
2. J. Sandor: Uber die Folge der Primzahlen. Mathematica (Cluj) 30 (53) (1988), 67–74.
´     ¨
3. L. Posa: Uber eine Eigenschaft der Primzahlen (Hungarian). Mat. Lapok 11 (1960),
124–129.
e
4. G. Robin: Estimation de la fonction de Tschebyshev θ sur le k-i`me nombre premier
et grandes valeurs de la fonction ω(n), nombre des diviseurs premier de n. Acta. Arith.
43 (1983), 367–389.
5. J. B. Rosser, L. Schoenfeld: Approximate formulas for some functions of prime
numbers. Illinois J. Math. 6 (1962), 64–89.

a
Facultatea de Mathematic˘,                                        (Received August 3, 1998)
s
Universitatea Bucure¸ti,
Str. Academiei nr. 14,
RO–70109 Bucharest 1, Romania

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