Consider a sequence of real numbers $\{a_n\}_n$. Let $\{a_{n_j}\}_j$ be a subsequence of $\{a_n\}_n$. Suppose I have shown that $\limsup_{j \rightarrow \infty} a_{n_j}=L$ with $|L|<\infty$. Is it true that $\limsup_{n \rightarrow \infty} a_{n}\geq L$?
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3Yes. ${}{}{}{}$ – Jan 25 '16 at 20:05
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You can proof this by contradiction if you want to check it rigorously. – noctusraid Jan 25 '16 at 20:06
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By definition $\limsup_{n\to\infty}\, a_n$ is the greatest number that you can achieve with a convergent subsequence of $\{a_n\}$. Since $\{a_{n_j}\}$ is a subsequence of $\{a_n\}$ we have that every subsequence of $\{a_{n_j}\}$ is a subsequence of $\{a_n\}$. Therefore $$\limsup_{n\to\infty}\, a_n\geq \limsup_{n\to\infty}\, a_{n_j}$$ because of the $\sup$'s properties.
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This characterization is not always taken as the definition, one can even guess that it was not for the OP if they are asking this question (but who knows). Anyway, please use
\limsup_{n\to\infty}. – Did Jan 25 '16 at 20:18