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Plase explain with an example the form correct of: $ \displaystyle {\limsup_{n\to \infty} s_n = s^* }$ with Rudin definition, because $s^* =\sup E $ and $ E $ is all point of subsequential limits. Because iam confused.

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Consider the sequence $$ s_n=(-1)^n\left(1+\frac{1}{n}\right). $$ Then the set of subsequential limit points is $E=\{-1,1\}$, and we have $$ \limsup_{n\to\infty}s_n=\sup E=1. $$

For a slightly more complicated example, look at the sequence which is something like $$ \{s_n\}=(1,1,\frac{1}{2},1,\frac{1}{2},\frac{1}{3},1,\frac{1}{2},\frac{1}{3},\frac{1}{4},1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},1,\ldots) $$ Then the set of subsequential limit points is $$ E=\left\{\frac{1}{n}:n\in\mathbb{N}\right\}, $$ and $$ \limsup_{n\to\infty}s_n=\sup E=1. $$

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  • $\begingroup$ Well more accurately $\limsup_{n\to \infty} \frac{1}{(n!)^{1/n}}=\sup E$, and $E=\{0\}$, but yes its correct. $\endgroup$ – Aweygan Aug 5 '16 at 17:51
  • $\begingroup$ Thanks! Now I can cocontinue $\endgroup$ – Jackson Aug 5 '16 at 18:32

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