Show that if all convergent subsequences of a sequence {sn} converge to 0 and {sn} is bounded, then {sn} converges to 0. My idea is to prove the contrapositive "If {sn}  does not converges to 0, then either {sn} is unbounded or there exists a subsequence that does not converge to 0"
My proof goes as follows:
Suppose sn is a sequence that does not converge to 0. Let {sn_k} be a subsequence of {sn}.
suppose to the contrary that {sn_k} converges to 0.
let ε > 0. Choose K natural number such that,
|sn_k| < ε for all k >= K
If n is at least K => |sn| < ε for all n >= K. Contradiction!
So, it must be that {sn_k} does not converge to 0.
°The proof seems incomplete but I can't think of a different way to tackle it
°Do I need to prove the part that says sn is unbounded? even if sn does not converge to zero it could still converge to another number in R
 A: Suppose $s_n$ is bounded and all it's subsequences which converge, converge to zero. We know that by the Bolzano Weierstrass theorem, at least one such subsequence exists.
I want to illustrate something interesting about the conditions that you have. Let $s_{n_k}$ be a subsequence of $s_n$. Then, $s_{n_k}$ is also bounded, and every subsequence of $s_{n_k}$, is a subsequence of $s_n$ anyway, so  also converges to zero. That is to say, every subsequence of $s_n$ has the property which $s_n$ has. I rewrite this in a fashion more revealing:

Every subsequence of $s_n$ has a convergent subsequence, converging to $0$.

This claim is enough to prove that $s_n \to 0$.This, we can do by contradiction: Suppose that $s_n \not \to 0$, then there exists $\epsilon>0$ and a subsequence $s_{n_k}$ such that $|s_{n_k}| > \epsilon$ for all $k$ (to see this statement, negate the definition of limit). But then, $s_{n_k}$ has a convergent subsequence $s_{n_{k_l}}$, which will converge to zero, by the statement in the yellow box. This cannot always stay $\epsilon$ far away from zero, right? So that is the contradiction.
That proves both your statement, and the statement in the yellow box, which is a more nicer looking statement.
A: Consider all terms of $a_n$ lying outside of $(-\epsilon,\epsilon)$. If there are infinitely many then Bolzano-Weierstrauss gives a convergent subsequence which we see does not converge to $0$. Contradiction, so there are only finitely many terms outside this interval and the sequence converges.
A: Assume $s_n$ does not converge to zero. By the definition of convergence:
$$\exists \epsilon>0\ \forall N\in\mathbb N\ \exists n>N \text{ such that }|s_n|\ge\epsilon$$
So you can construct recursively a subsequence $s_{n_k}$ with $|s_{n_k}|\ge\epsilon$
If $s_n$ is unbounded, then we are done.
If $s_n$ is bounded, then $s_{n_k}$ is bounded, thus it has a convergent subsequence $s_{n_{k_p}}\to l$, so $|s_{n_{k_p}}|\to |l|\ge\epsilon$. So $s_{n_{k_p}}$ has a non-zero limit, which was to be proved.
A: $\lim_{n\to \infty}s_n=0$ iff for all $r>0$ the set $V(r)=\{n:|s_n|>r\}$ is finite (i.e. bounded above).
So if $(s_n)_n$ does not converge to  $0$ then there exists $r>0$ such that $V(r)=\{n:|s_n|>r\}$ is infinite, so for such an $r,$ let $V(r)=\{f(n)\}_n$ where $f$ is strictly increasing. Then $(s_{f(n)})_n$ is a subsequence not converging to $0$ because $|s_{f(n)}|>r$ for all $n.$ 
