When does a limit of a sequence equal both its lim sup and lim inf? 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!
 A: 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.
A: 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 :)
