Polynomial-Exponential Inequality

Question

Consider the following sequences $$ x_n = \sup \{ k \in \mathbb{N} : e^{2^k} 2^k \leq n \} $$ $$ y_n = \sup \{ k \in \mathbb{N} : e^{2^k} 2^{2k} \leq n \} $$ Clearly $x_n \geq y_n$ but I would like to show that infact $x_n=y_n$ for all sufficiently large $n$. Is there a simple way to do this?

Answer 1

This is blatantly and obviously false. (It may be asymptotically true in some sense, but that's another story.)

Say, $a_k=e^{2^k} 2^k$, and $b_k=e^{2^k} 2^{2k}$. Now it is rather obvious that $a_1<b_1<a_2<b_2<a_3<b_3\dots$, where every "<" contains a pretty wide interval including many natural numbers. Well, for every $n$ such that $b_k<n<a_{k+1}$ you will have $x_n=y_n=k$, but then there are many other numbers such that $a_k<n<b_k$, and for them it will be $x_n=k,\;y_n=k-1$.