Dependencies in 'epsilon of room'-style proofs

Question

I want to try and prove that

If $\{a_n\}_{n=1}^\infty$ is a positive sequence converging to zero, then there exists $N \in \mathbb{N}$ such that $a_{n+N} \leq a_n$ for every $n \in \mathbb{N}$.

Proof. Given $\varepsilon > 0$, it was easy to prove that there exists $N$ that depends on $\varepsilon$, such that $a_{n+N} \leq a_n + \varepsilon$. For as $\{\sup_{n \geq k}a_n\}_{k=1}^\infty$ is decreasing, I pick $N$ such that $\sup_{n \geq k} a_n < \varepsilon$ whenever $k \geq N$. This means that for every $n \in \mathbb{N}$, $a_{n+N} < \varepsilon \leq \varepsilon + a_n$.

Question: I have heard of 'Epsilon of room'-proofs, but I cannot find out whether I am allowed to let $N$ depend on $\varepsilon$ in such proofs, or not. Could you provide and answer to whether such a dependecy is allowed, and also explain what kind of dependencies are allowed and forbidden in such proofs.

Answer 1

What you want to prove is not true. For example, define a sequence $(a_n)_{n \geq 1}$ by $$\forall n \geq 1 \text{ such that there exists }k \text{ with } 2^k < n < 2^{k+1}, \quad a_n = \frac{1}{k+1}$$ and $$a_{2^k} = \frac{2}{k+1}$$

For such a sequence, there is no such $N$.

Answer 2

$\newcommand{\eps}{\varepsilon}$If the statement you are trying to prove is of the form "for all $\eps > 0$, there exists $N\in \mathbb{N}$ such that...", then $N$ is allowed to depend on $\eps$.

If however it was "there exists $N\in\mathbb{N}$ such that for all $\eps > 0$ ...", then $N$ would not be allowed to depend on $\eps$.

In summary, it depends on the order of the "there exists" and the "for all" parts of the statement. In general, a variable $N$ in a "there exists" part of a statement is allowed to depend on any variables that came in "for all" parts before $N$ in the statement, but not on variables that come after $N$ in the statement.

Answer 3

The sequence

$$1,2,1/2,1/2,1,1/3,1/3,1/3,2/3,1/4,1/4,1/4,1/4,2/4,\dots $$

shows the result fails.