4
$\begingroup$

Mills proves that there is a real number $A$ such that $\lfloor A^{3^n}\rfloor$ is a prime number for every integer $n\ge 1$.

After having proved that there is a prime between any two sufficiently large consecutive cubes, Mills constructs an infinite sequence of primes $P_0,P_1,P_2,$ $\cdots$ such that $(P_n)^3<P_{n+1}<(P_n+1)^3$.

He then defines two functions that he applies to his sequence of primes $P_0,P_1,P_2,$ $\cdots$. Such functions are:

$$u(n)=\sqrt[3^n]{P_n}$$ $$v(n)=\sqrt[3^n]{P_n+1}$$

He then proves the following three statements:

1) $$\sqrt[3^n]{P_n+1}>\sqrt[3^n]{P_n}$$

2) $$\sqrt[3^{n+1}]{P_{n+1}}>\sqrt[3^n]{P_n}$$

3) $$\sqrt[3^{n+1}]{P_{n+1}+1}<\sqrt[3^n]{P_n+1}$$

Unfortunately, I don't understand what follows, starting from "It follows at once that the $u_n$ form a bounded monotone increasing sequence". What is a bounded monotone increasing sequence? And what is $\lim_{n\to\infty}{u_n}$? I'd like to understand this final part of the proof.

Thanks in advance!!

$\endgroup$
2
4
+150
$\begingroup$

In general, a sequence $a_1, a_2, \ldots$ is called monotone increasing if it satisfies $a_{n+1} > a_n$ for every $n$. In other words, every element of the sequence is strictly greater than the previous element. In his case, equation (2) in your question expresses precisely that $u_{n+1} > u_n$ for every $n$, so indeed, the sequence $u_1, u_2, \ldots$ is monotone increasing.

On the other hand, a sequence $a_1, a_2, \ldots$ is bounded if there are constants $C$ and $D$ such that $C \leq a_n \leq D$ for every $n$. In other words, every element of the sequence lies between $C$ and $D$ on the number line. We call $C$ a lower bound for the sequence, and $D$ an upper bound. In this case, we can set $C = u_1$ and $D = v_1$, and then we claim that $C \leq u_1 \leq D$ is true for all $n$. To see this, let $n$ be any natural number. From equation (2) in your post, we see that $u_1 < u_2 < \ldots < u_n$, and similarly we see from equation (3) that $v_1 > v_2 > \ldots > v_n$. From equation (1), we have $u_n < v_n$. Putting all these equations together and using our definitions of $C$ and $D$, we find that $$C = u_1 < u_n < v_n < v_1 = D,$$ so in particular we have $C \leq u_n \leq D$ for every $n$. Therefore we conclude that the sequence $u_1, u_2, \ldots$ is bounded, as claimed.

If $a_1, a_2, \ldots$ is a sequence, and $A$ a number, then we say that $A$ is the limit of the sequence if, roughly stated, the numbers $a_n$ get closer and closer to $A$ when $n$ becomes bigger and bigger (the precise definition of limits is a bit more technical that this, but this is the idea). For example, the sequence $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$ has 0 as limit, because the terms get very close to 0 as we progress in the sequence. As it happens, not every sequence has a limit. For instance, the sequence $1, 0, 1, 0, 1, 0, \ldots$ does not get closer and closer to any number. But if the sequence $a_1, a_2, \ldots$ has a limit, then this limit is unique, and we denote this limit by $\lim_{n \to \infty} a_n$.

Now, it is an important fact from real analysis, called the monotone convergence theorem, that if a sequence is both bounded and monotone increasing, then it must have a limit, and this limit is in fact equal to the smallest real number that is an upper bound for the sequence. In our case, we know that $u_1, u_2, \ldots$ is both bounded and monotone increasing, so we can conclude that $u_1, u_2, \ldots$ has a limit $\lim_{n \to \infty} u_n$.

$\endgroup$
1
  • 2
    $\begingroup$ Thank you! Your answer helped me to understand Mills' proof. $\endgroup$
    – User X
    Oct 25 '15 at 23:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.