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Let $\{a_n\}_{n \in \mathbb{N}} \in \mathbb{R_+}$ a sequence of real positive numbers such that $$ \lim_{n \to \infty} a_n = 0 $$

I would like to know if is true that $\exists m \in \mathbb{N} : a_n \leq a_m \forall n \in \mathbb{N} $

Thanks

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    $\begingroup$ It better exists, otherwise the sequence is unbounded above ... Of course, this also follows at once from the very definition of limit. $\endgroup$ – DonAntonio Mar 2 '17 at 23:14
  • $\begingroup$ @DonAntonio can you write a proof ? $\endgroup$ – Matey Math Mar 2 '17 at 23:18
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Let $\varepsilon = a_1 >0$.

By convergence $a_n \to 0$, there exists a threshold $N$ after which all indices $n \geqslant N$ satisfy $a_n < \varepsilon$.

Furthermore, there exists a $m \in \{1,...,N-1\}$ such that $a_m = \max \{a_1,...,a_{N-1}\}$.

Now, given any positive integer $n$, we either have

$n \geqslant N\quad \Longrightarrow \quad a_n < \varepsilon = a_1 \leqslant a_m$,

or we have

$n < N \quad \Longrightarrow \quad a_n \in \{a_1,...,a_{N-1}\} \quad \Longrightarrow \quad a_n \leqslant a_m$.

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The statement is true, proof:

Take $\epsilon > 0$, then by convergence there exists an $N$ s.t. $a_n < \epsilon$ for $n > N$. Let $m = \text{argmax}_{n \leq N} a_n$, if $a_m > \epsilon$ it follows that $a_m > a_n \; \forall n \in \mathbb{N}$.

For $\epsilon$ small enough the condition $a_m > \epsilon$ must be satisfied, unless the whole sequence is $0$, in which case the proposition also holds.

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Since the sequence approaches $0$, it must be bounded, hence it must have a least upper bound, necessarily positive. If the least upper bound is not in the sequence, there must be an infinite subsequence which converges to it, contradiction, since every infinite subsequence must converge to $0$ (given that the main sequence converges to $0$). It follows that the least upper bound is an element of the sequence, so there is a maximum term, which is what was to be proved.

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