Dependencies in 'epsilon of room'-style proofs 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 $\varepsilon &gt; 0$, 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 $\sup_{n \geq k} a_n &lt; \varepsilon$ whenever $k \geq N$. This means that for every $n \in \mathbb{N}$, $a_{n+N} &lt; \varepsilon \leq \varepsilon + a_n$.

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

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