# Show $(a_n)_{n=1}^\infty$ converges to $L \in \mathbb{R}$ iff the limit superior and limit inferior converge to $L$

Show $$(a_n)_{n=1}^\infty$$ converges to $$L \in \mathbb{R} \iff lim\ sup\ a_n = lim\ inf\ a_n = L$$.

Direction: $$\implies$$ Suppose $$(a_n)$$ converges to $$L \in \mathbb{R}$$, show $$lim\ sup\ a_n = lim\ inf\ a_n = L$$.

We know since $$a_n$$ converges, we have that $$\forall \epsilon > 0, \exists N, n \geq N \implies |a_n - L| < \epsilon$$.

Note that we know $$lim\ sup\ a_n$$ and $$lim\ inf\ a_n$$ converges (note $$a_n$$ is bounded since it is convergent and sequences $$\sup\{a_k\ |\ k \geq n\}\ and\ \inf\{a_k\ |\ k \geq n\}$$ are monotone).

We seek to show:

1. $$\forall \epsilon > 0, \exists N',\ n \geq N' \implies |sup\ a_n - L| < \epsilon$$
2. $$\forall \epsilon > 0, \exists N'',\ n \geq N'' \implies |inf\ a_n - L| < \epsilon$$

Consider the set $$\sup\{a_k\ |\ k \geq n\}$$ where $$n \geq N$$. We know since $$\forall \epsilon > 0, |a_n - L| < \epsilon$$, we have $$|\sup\{a_k\ |\ k \geq n\} - L| < \epsilon$$. Analogously for $$|inf\ a_n - L|$$.

So set $$N' = N'' = N$$, we have that 1 and 2 hold and so the limit superior and inferior converge to $$L$$.

Direction: $$\impliedby$$ Suppose $$lim\ sup\ a_n = lim\ inf\ a_n = L$$...

By the uniqueness of limits, we know $$\sup\{a_k\ |\ k \geq n\} = \inf\{a_k\ |\ k \geq n\}$$. We know since the supremum and the infimum are equal, $$|\{a_k\ |\ k \geq n\}| = 1$$.

So let $$N = n$$. $$\forall \epsilon > 0, \ m \geq N \implies |a_m - L| < \epsilon$$. And so $$(a_n)_{n=1}^\infty$$ converges to $$L$$.

Apologies if the notation is confusing. Want to know if my proof makes sense and if not where it needs fixing / a better way to prove this.

It's fine but you need to prove it effectively! Here's my opinion:

Actually, if $$y_n=\text{sup}\;\{a_k: k \geq n\}$$ and if $$\lim y_n$$ exist, then we say $$\lim y_n=\limsup a_n$$

We already know $$y_n$$ converges, so in this place, we assume $$y_n \rightarrow y$$ ,say

Now, $$a_n \rightarrow l$$ means $$l-\varepsilon$$ and $$l+\varepsilon$$ are lower and upper bounds for $$\{a_n,a_{n+1},\cdots\}$$

It follows that $$\vert y_n -l \vert < \varepsilon$$ for $$n$$ large. It means $$l-\varepsilon < y_n for $$n$$ large.

Consequently$$l-\varepsilon < y for $$n$$ large.

Since $$\varepsilon$$ is arbitrary, $$y=l$$

The case for $$\text{inf}\{a_k\}$$ is analogous!