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 $$l-\varepsilon < y_n <l+\varepsilon$$ for $n$ large.

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

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

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


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