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Problem: Show that if all convergent sub-sequences of a sequence ${s_n}$ converge to $0$ and ${s_n}$ is bounded, then ${s_n}$ converges to $0$.

Attempt: I am trying to solve this problem using the contra-positive. Here is my attempt so far. If $s_n$ diverges to positive or negative infinity, then ${s_n}$ is not bounded. If $s_n$ converges to a non-zero real number, then any sub-sequence of $s_n$ converges to the same value (which is not zero).

Now where I am stuck is the case where $s_n$ diverges. Any hints much appreciated.

Edit: This part of the question is important: "all convergent sub-sequences of a sequence ${s_n}$ converge to $0$ ", this does not assume that any sub-sequence is in fact convergent.

Also I believe the contrapositive of this statement is that $s_n$ does not converge to zero implies that there exists a sub-sequence of $s_n$ that does not converge to $0$ or that $s_n$ is un-bounded.

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  • $\begingroup$ Can you prove it knowing that $s_{2n}$ and $s_{2n+1}$ are convergent? $\endgroup$ – arberavdullahu Jan 29 '17 at 19:30
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    $\begingroup$ How do you "know" that $s_{2n}$ and $s_{2n+1}$ are convergent? The problem doesn't say that. $\endgroup$ – zipirovich Jan 29 '17 at 19:35
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If the sequence $(s_n)_{n\geq1}$ does not converge to $0$ then there is an $\epsilon_0>0$ such that for any $n\geq1$ there is a $k>n$ with $|s_k|\geq\epsilon_0$. This allows to select a strictly increasing sequence $l\mapsto k_l\in{\mathbb N}$ such that $|s_{k_l}|\geq\epsilon_0$ for all $l\geq1$, hence $\epsilon_0\leq|s_{k_l}|\leq C$ for all $l\geq1$, where $C$ is the assumed overall bound for the $|s_n|$. By Bolzano's theorem the sequence $\bigl(s_{k_l}\bigr)_{l\geq1}$ has a convergent subsequence. Its limit point is necessarily $\ne0$, contrary to the assumption about the original sequence $(s_n)_{n\geq0}$.

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  • $\begingroup$ Why are you assuming that $s_n$ is bounded? $\endgroup$ – IntegrateThis Jan 29 '17 at 19:52
  • $\begingroup$ You say so in the title of the question. $\endgroup$ – Christian Blatter Jan 29 '17 at 19:54
  • $\begingroup$ Sorry It's just that the contrapositive would be to show that there exists a sub-sequence that is convergent not converging to 0, or the sequence is not bounded. $\endgroup$ – IntegrateThis Jan 29 '17 at 19:55
  • $\begingroup$ The assumption in the contrapositive is that $s_n$ does not converge to 0. $\endgroup$ – IntegrateThis Jan 29 '17 at 19:57
  • $\begingroup$ But I believe I can modify my proof to include yours with 2 more cases thanks. $\endgroup$ – IntegrateThis Jan 29 '17 at 20:05
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We know that there exist a convergent subsequence say $s_{m_i}$ such that $0=\lim s_{m_i}=\lim sup s_n$, similiar that exist a convergent subsequence say $s_{n_i}$ such that $0=\lim s_{n_i}=\lim inf s_n$
Knowing that $$inf s_n\leq s_n\leq sup s_n$$ we can use Squeeze lemma to show that $\lim s_n=0$

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Yes, we can prove it by contrapositive (or by contradiction? I always confuse the two). We need to show that $\lim\limits_{n\to\infty}s_n=0$, which means $$\forall \varepsilon>0 \; \exists N\in\mathbb{N} \; \forall n\ge N \; \colon \; |s_n|<\varepsilon.$$ Assume $\lim\limits_{n\to\infty}s_n=0$ is not true, the the negation of the definition says that $$\exists \varepsilon>0 \; \forall N\in\mathbb{N} \; \exists n\ge N \; \colon \; |s_n|\ge\varepsilon.$$ Fix this $\varepsilon$. For each $N\in\mathbb{N}$, there exists some $n=n_N\ge N$ such that $|s_{n_N}|\ge\varepsilon$. So we have an infinite subset $\{s_{n_1},s_{n_2},s_{n_3},\ldots\}$ of the given sequence. Since the original sequence is bounded, this subset is bounded too, and therefore it has a limit point. That means that there's a subsequence converging to that limit point. But since all elements are bounded away from $0$, we've constructed a subsequence that converges, but not to $0$, contradicting the given condition.

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