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?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.