Bertrand’s conjecture, proof of
This is a version of Erdős’s proof as it appears in Hardy and Wright.
We begin with the following lemma.
Let be a positive integer and be a prime. The highest power of dividing is , where is the floor of .
Let divide with as large as possible. Every th term of the sequence is divisible by , so contributes a factor to . There are such factors. Every th term contributes an extra factor above that, providing new factors. In general, the th terms contribute extra factors to . So the highest power of dividing is . ∎
We now prove the theorem.
Let be a minimal counterexample to the claim. Thus there is no prime such that .
In the sequence of primes
each succeeding term is smaller than the double of its predecessor. This shows that .
The binomial expansion (http://planetmath.org/BinomialTheorem) of has terms and has largest term . Hence
For a prime define to be the highest power of dividing . To compute , we apply the lemma to and . We get that
The terms of the sum are all 0 or 1 and vanish when , so , that is, .
For , and so for we can apply the previous formula for and find that it is zero. So for all , the contribution of the primes larger than is zero.
If , all the terms for higher powers of vanish and . Since is at most 1, an upper bound for the contribution for the primes between and is the product of all primes smaller than . This product is , where is the Chebyshev function
There are at most primes smaller than and by the inequality their product is less than . Combining this information, we get the inequality
Taking logarithms and applying the upper bound of for (http://planetmath.org/UpperBoundOnVarthetan), we obtain the inequality , which is false for sufficiently large , say . This shows that .
Since the conditions and are incompatible, there are no counterexamples to the claim. ∎
- 1 G.H. Hardy, E.M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, 1938.