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I need to prove that $\lim\sup(|s_n|) = 0$ if and only if $\lim(s_n) = 0$. I totally get the intuition behind this implication but not sure how to give a formal proof of it.

Here is my intuition: for the forward direction, $\lim\sup(|s_n|) = 0$ means that the magnitude of each term is getting arbitrarily small. Hence, $\lim(s_n) = 0$. For the reverse direction, if $\lim(s_n) = 0$ then for any subsequence $\left\{ s_n \right\}_{n > N}$ for some $N > 0$, it is clear that $\sup(|s_n|) \to 0$.

Any suggestions on how to proceed with the formal proof?

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  • $\begingroup$ my intuition is that $\liminf |s_n| = \limsup |s_n| = 0$, hence $\lim |s_n| = 0$, hence $\lim s_n = 0$; or if you want to prove directly, just write down the definition of $\limsup$. $\endgroup$ – Du Phan Mar 12 '14 at 9:16
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Suppose $\lim\sup|s_n|=0$. Take $\varepsilon>0$. There exists $n_0\in\mathbb N$ such that $|s_n|<\varepsilon$ for $n\geq n_0$, that is, $-\varepsilon<s_n<\varepsilon$, so $\lim s_n=0$.

The other implication is proven the same way.

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