Let us consider a countable set of sequences of positive numbers $\{(x\_n^{(1)}),(x\_n^{(2)}),\dots\}$ for which we have


* $(\forall k\in\mathbb{N}) \ \lim\_n x\_n^{(k)} = +\infty$, and
* $(\forall k\in\mathbb{N}) \ \lim\_n\frac{x\_n^{(k+1)}}{x\_n^{(k)}}=0$ (the growth rate of the sequences is in the descending order).


My question is the existence of a minimal element, i.e. is there a sequence of positive numbers $(y\_n)$ such that $\lim\_n y\_n=+\infty$ and for every $k\in\mathbb{N}$ we have $\lim\_n\frac{y\_n}{x\_n^{(k)}}=0$?


Of course, any bounded sequence satisfies the second condition, but that is not satisfactory to me. 


An example of the above situation where the existence of $(y\_n)$ is obvious is $x\_n^{(k)}=n^{\frac{1}{k}}$, where we can take $y\_n=\ln n$. 



The minimal growth rate of the countable family of sequences
Joel's answer gives the history of the result, and Fan Zheng's comment gives a one-word proof, "diagonalization", but it might be worthwhile to make the argument more explicit. The strategy for defining the sequence $(y\_n)$ is to make it a sequence of integers that is monotone non-decreasing but grows very slowly. That is, $y\_n$ will be 1 for $1\leq n<j\_1$, and then it will be 2 for $j\_1\leq n<j\_2$, then 3 for $j\_2\leq n<j\_3$, and so forth. The intervals of constancy, $[j\_i,j\_{i+1})$, will be chosen to be quite long, so that the sequence $(y\_n)$ increases very slowly, but notice that this sequence will tend to infinity. What remains is to make sure that $y\_n$ is o$(x^{(k)}\_n)$ for every $k$, and that will be achieved by choosing the $j$'s large enough. Specifically, $j\_m$, the place where the $(y\_n)$ sequence jumps from $m$ to $m+1$, should be chosen to be a number $j$ so large that, for all $k\leq m$ and all $n\geq j$, $x^{(k)}\_n>(m+1)^2$. Of course, such a $j$ exists because each of the $x^{(k)}$ sequences tends to infinity. (The square of $(m+1)$ here could be replaced with any other exponent $>1$, or indeed with anything that ensures that the value of $x^{(k)}\_n$ is way bigger than the new value $m+1$ in the $y$-sequence.) What this choice of the $j\_m$'s achieves is that, for any fixed $k$, you'll have $y\_n$ way smaller than $x^{(k)}\_n$ once $n$ is past $j\_k$. Therefore, $y\_n$ is o$(x^{(k)}\_n)$, as required.


