# Understanding the final part of the proof of Mills' theorem

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.

• Oct 19 '15 at 20:57
• Sorry, I wanted to say "It'd be great if I could have an answer with all the necessary details!". Oct 21 '15 at 0:41

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 $\lim_{n \to \infty} u_n$.