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Let $\{x_n\}_{n\in\mathbb{N}}\subseteq\overline{\mathbb{R}}$ a sequence. On some texts the definition of $\text{limsup}$ is as follows:

Definition 1. $$\text{limsup}x_n=\inf_{k\ge1}\bigg(\sup_{n\ge k}x_n\bigg)$$

while other texts other texts give the following definition

Definition 2. $$\text{limsup}x_n=\inf_{k\ge 0}\bigg(\sup_{n\ge k}x_n\bigg)$$

Question Are the two definitions equivalent?

Thanks!

Jack J.
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  • It all depends on what is the starting value of $n$ (or $k$), that is whether $N={0,1,2,\ldots}$ or ${1,2,3,\ldots}$. – Chrystomath Feb 19 '19 at 12:50

1 Answers1

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As mentioned in the comment, the definition depends on whether your definition of $\Bbb N$ includes zero or not, i.e. if the first index of a sequence is $0$ or $1$. But it makes no difference even if $\Bbb N = \{ 0,1,2,\ldots \}$: For each $k$ is $$ \sup_{n\ge k+1}x_n \le \sup_{n\ge k}x_n $$ because the supremum of a smaller set is taken on the left-hand side. In other words, the sequence $$ \bigg(\sup_{n\ge k}x_n\bigg)_k $$ is decreasing, and it's infimum (which is equal to its limit for $k\to \infty$) does not depend on whether the term for $k=0$ is included or not.

Martin R
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