## Firoozbakht QED?

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• The Firoozbakht s conjecture (1982) is equal to: (p_(n+1))^(n)
Message 1 of 2 , Apr 8, 2014

The Firoozbakht's conjecture (1982) is equal to:

(p_(n+1))^(n) < (p_n)^(n+1).

Then the natual log is:

n*ln(p_(n+1)) < (n+1)*ln(p_n).

Now,
ln(p_n) <= ln(n) + ln(ln(n)) + 1, for n >= 2. (Dusart 2010)

And, because p_n >= n*ln(n), for n >= 2;
(Dusart 1999)
the nat log of p_(n+1) is:
ln(n+1) + ln(ln(n+1)) <= ln(p_(n+1)), for n >= 2.

So,
n*(ln(n+1) + ln(ln(n+1))) < (n+1)*(ln(n) + ln(ln(n)) + 1).
Divided by n*(ln(n) + ln(ln(n)) + 1):
(ln(n+1) + ln(ln(n+1)))/(ln(n) + ln(ln(n)) + 1) < (n+1)/n

This inequality is true because the left-side increases slower than the right-side.

Is this a QED?

John W. Nicholson
• ... No. You failed to understand the concept of inequality. Let A(n)=(n+1)/n; B(n)=log(prime(n+1))/log(prime(n));
Message 2 of 2 , Apr 10, 2014

> Is this a QED?

No. You failed to understand the concept of inequality. Let

A(n)=(n+1)/n;
B(n)=log(prime(n+1))/log(prime(n));
C(n)=(log(n+1)+log(log(n+1)))/(log(n)+log(log(n))+1);

For n > 1, it is easy to prove that
A(n) > C(n) and
B(n) > C(n).

At elementary school you should have learnt that those two
inequalities do not determine which of A(n) and B(n) is the
larger.

David
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