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!

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.