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This question already has an answer here:

If $\{a_{n}\}$ is a sequence of real numbers, prove that $\displaystyle\limsup_{n \to \infty}{a_n} = \displaystyle\lim_{N \to \infty} \sup{\{a_n: n \ge N\}}$.

It is sort of intuitively clear, since $\sup_{n > 1}{a_{n}}$, $\sup_{n > 2}{a_{n}}$ ... is nonincreasing sequence and for any N, we have that $\sup_{n\ge N} a_n \le \sup_{n\ge 1}a_n$. So eventually $\sup_{n\ge N} a_n$ should be equal to $\displaystyle\limsup_{n \to \infty}{a_n}$, but how to prove it formally? By the definition of limsup from Rudin, limsup is a supremum of a set of subsequential limits, but it does no seem to help here.

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marked as duplicate by Brahadeesh, Paul Frost, Namaste, Sou, Andrei Nov 18 '18 at 20:26

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You could try to construct a subsequence with $$a_{k_n}\ge\sup_{m>n}a_m-2^{-n}.$$

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  • $\begingroup$ Sorry, I do not understand. How can it help me? $\endgroup$ – user0347284 Nov 27 '16 at 0:57
  • $\begingroup$ If successful, you get a sub-sequence that converges to $\lim_n\sup_{m>n}a_n$. $\endgroup$ – Dr. Lutz Lehmann Nov 27 '16 at 1:31
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I try to give an answer which is independent on the various definitions of $\limsup a_n$, basing on the two following elementary lemmas.

Lemma 1. Let $\langle a_n \rangle = \{ a_n : n\in\mathbb{N}\}$ be a (not necessarily strictly) monotonically decreasing succession of extended real numbers (i.e. $a_n\in\overline{\mathbb{R}}= \mathbb{R}\cup\{+\infty, -\infty\}$ for each $n\in\mathbb{N}$). Then $$ \lim_{n\to\infty} a_n = \inf\langle a_n \rangle $$ Proof. The case $\inf\langle a_n \rangle=\infty$ is trivial. If $\inf\langle a_n \rangle\in\mathbb{R}\cup\{-\infty\}$, from the definition of infimum of a set of real numbers we have that, for any given real number $\varepsilon > 0$, there exists a natural number $m$ $$ \begin{cases} a_m < -\varepsilon & \text{if } \inf\langle a_n \rangle =-\infty,\\ 0 \leq a_m - \inf\langle a_n \rangle < \varepsilon & \text{if } \inf\langle a_n \rangle \in \mathbb{R}. \end{cases} $$ But since $\langle a_n \rangle$ is monotonically decreasing, for every natural $n\geq m$ $$ \begin{cases} a_n < -\varepsilon & \text{if } \inf\langle a_n \rangle =-\infty,\\ 0 \leq a_n - \inf\langle a_n \rangle < \varepsilon & \text{if } \inf\langle a_n \rangle \in \mathbb{R}. \end{cases} \Longleftrightarrow \lim_{n\in\mathbb{N}} a_n = \inf\langle a_n \rangle.\quad\blacksquare $$

Lemma 2. Let $\langle a_n \rangle = \{ a_n : n\in\mathbb{N}\}$ be a (not necessarily strictly) monotonically decreasing succession of extended real numbers, and let $\langle a_n \rangle_{n\geq{N}} = \{ a_n : n\geq N, n\in\mathbb{N}\}$ a subsequence obtained from $\langle a_n \rangle$ by omitting its first $N$ terms. Then $$ \inf\langle a_n \rangle = \inf\langle a_n \rangle_{n\geq{N}} $$ Proof. It is a trivial consequence of the definition of infimum of a set of real numbers and of monotone sequence of real numbers. $\blacksquare$

Put $A_n = \sup \{ a_m : m\geq n\}$ for each $n\in\mathbb{N}$: now $\langle A_n \rangle$ is a monotonically decreasing sequence of extended real numbers, as noted by user0347284: therefore, by applying to it the two preceding lemmas one gets the sought for proof: $$ \limsup_{n \to \infty} a_n = \inf\langle A_n \rangle = \inf\langle A_n \rangle_{n\geq{N}} = \lim_{N \to \infty} A_N = \lim_{N \to \infty} \sup \{ a_n : n\geq N\} $$

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