# limit supreme and infimo [duplicate]

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.

## 1 Answer

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 $$\limsup_{n\to\infty}s_n=\sup E=1.$$

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