Where ? and ? are constants (0 < ? < ?) to show that there exists a constant ? > 1 such that for every ? ≥ 2 there is a prime satisfying ? < ? ≤ ??.
Solution.
Let ? be any number such that ? > ?/?. Then
∑ log ?
?<?≤??
= ∑ log ?
?≤??
− ∑ log ?
?≤?
≥ ?(??)− ?? > ?? − ?? = 0
Since this sum is non-zero, there exists a prime ? satisfying ? < ? ≤ ??.
Let us show now that ? = ?/? also satisfies the condition. Fix any ? ≥ 2.
As it is proved, the interval (?, (1 + ?/?)?] contains at least one prime number. Let ? be the
minimal prime in this interval. Suppose ? > (?/?)?. Then
? = 1
2
(
?
?
+
?
?
) > 1
2
(
?
?
+
?
?
) = ?
?
and ? = 1
2
(
?
?
+
?
?
) < 1
2
(
?
?
+
?
?
) = ?
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