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Suppose $(a_n)_{n\ge1}$ is a positive (so non-zero), real sequence with $\lim_{n\rightarrow\infty}a_n=L$, where $L\in[0,\infty]$. Is this equivalent with $\limsup_{n\rightarrow\infty}a_n=\liminf_{n\rightarrow\infty}a_n=L$? In particular, does this hold for the infinity case? If so, why? Thank you!

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  • $\begingroup$ In the case of infinity it doesn’t hold en.m.wikipedia.org/wiki/Limit_superior_and_limit_inferior (under properties) $\endgroup$
    – gbox
    Oct 16 '17 at 21:15
  • $\begingroup$ I think that it does hold for infinity, actually, by the definition of divergence to infinity. $\endgroup$
    – Michael L.
    Oct 16 '17 at 21:24
  • $\begingroup$ @gbox I don't see directly what you mean. Could you be more precise about the position of this fact in the page of the link you posted, or elaborate what you mean in an answer? $\endgroup$
    – Tyron
    Oct 16 '17 at 21:25
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    $\begingroup$ If $\lim_{n\to \infty} x_n = \infty$, then for all $M > 0$, there is an $n_M\in \mathbb{N}$ such that $x_n\geq M$ for all $n\geq n_M$. Letting $y_m := \inf_{n\geq m} x_n$, we have that for $M\geq 0$, $y_m\geq M$ for all $m\geq n_M$ due to the fact that $y_m$ is increasing in $m$. Therefore, $y_m\to \infty$ as $m\to \infty$. By the fact that $z_m := \sup_{n\geq m} x_n$ dominates $y_m$, we also have $z_m\to \infty$. $\endgroup$
    – Michael L.
    Oct 16 '17 at 21:28
  • $\begingroup$ @Tyron it's incorrect and therefore won't appear on the Wikipedia page. $\endgroup$
    – Michael L.
    Oct 16 '17 at 21:30
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It is equivalent. If $\limsup_{n \to \infty} a_n = \liminf_{n \to \infty} a_n = L \in \mathbb{R}$, then the set of partial limits will only contain $L$ (If it contains anything else: say bigger than $L$: then L is not the supremum of the set, and similarly for the other cases). Now, if $\lim_{n \to \infty} a_n \neq L$, then there exists $\varepsilon > 0$ s.t. for any $N$ there exists $k > N$ s.t. $|a_k - L| \geq \varepsilon$. The sequence $a_k$ (formalization needed here) contains a sequence that converges to a real number or plusminus infinity (any sequence does), and it cannot converge L (since $|a_k - L| \geq \varepsilon$), denoted by $K$. Then $K\neq L$ but K is a partial sum. Contradiction! Therefore $\lim_{n \to \infty} a_n = L$.

The other direction is clear: if a sequence is convergent, any of its subsequence will converge with her, and therefore the set of partial limits will contain its limit only.

I trust you to fill in the $+-\infty$ case :)

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The $\displaystyle\liminf_{n \to +\infty} a_n$ denotes the smallest adherence value of the sequence $(a_n)_n$. The $\displaystyle\limsup_{n \to +\infty} a_n$ denotes the greatest adherence value of the sequence $(a_n)_n$. Recall that $\ell$ is an adherence value of $(a_n)_n$ if there is a subsequence of $(a_n)_n$ which converges to $\ell$. Thus, if the limit inf is equal to the limit sup then the whole series converges to this common limit. The reverse is also true.

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  • $\begingroup$ Thank you. Can the same reasoning be applied to the case in which $L=\infty$, with $L$ as in my question? $\endgroup$
    – Tyron
    Oct 16 '17 at 21:26
  • $\begingroup$ In my answer, $\ell$ can be a real number or $\pm \infty$. $\endgroup$ Oct 17 '17 at 6:53

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