I have yet to see very straightforward proofs that convergence of a sequence implies equality of $\liminf$ and $\limsup$, so I'd like to attempt to present one here:

Statement: If $\{a_k\}$ converges, $$\liminf_{ k \to \infty} \{a_k\} = \limsup_{k \to \infty} \{a_k\}.$$

Proof: Let $\epsilon >0$. Recall that $$\liminf_{ k \to \infty} \{a_k\} = \lim_{n \to \infty} \inf_{k \geq n} \{a_k\},$$ since $\inf_{k \geq n} \{a_k\}$ is monotonically decreasing in $n$. Similarly for $\limsup$. Thus, there is $N_1$ so that if $n \geq N_1$,

$$\left|\liminf_{ k \to \infty} \{a_k\} - \inf_{k \geq n} \{a_k\}\right| < \frac{\epsilon}{4} \qquad(1)$$ and $$\left|\limsup_{ k \to \infty} \{a_k\} - \sup_{k \geq n} \{a_k\}\right| < \frac{\epsilon}{4} \qquad (2).$$

Now, since the sequence converges, it is cauchy. Let $N_2$ be such that for $n,m \geq N_2$, $$\left|a_n-a_m\right| < \frac{\epsilon}{8}.$$

Let $N = \max\{N_1,N_2\}$. By the definition of inf, there exists $a_j$ with $j \geq N$ so that $$\left|a_j - \inf_{k \geq N} \{a_k\}\right| < \frac{\epsilon}{8}$$ and so, for $n \geq N$,

$$\left|a_n - \inf_{k \geq N} \{a_k\}\right| < \frac{\epsilon}{4} \qquad (3)$$ by the triangle inequality, and the fact that the sequence is cauchy.

Similarly, for $n \geq N$,

$$\left|a_n - \sup_{k \geq N} \{a_k\}\right| < \frac{\epsilon}{4} \qquad (4).$$

Combining (1)-(4),

we obtain $$\left|\liminf_{ k \to \infty} \{a_k\} - \limsup_{ k \to \infty} \{a_k\} \right| < \epsilon$$ and since $\epsilon$ was arbitrary, we have our desired equality.

Indeed, although this proof may be longer than standard proofs, I believe it more accurately portrays to the students that we are trying to control oscillation after some index, similar to the oscillation of a continuous function i.e. show that this proof is essentially the discrete analogue of the proof that a function is continuous iff its oscillation tends to 0.

  • $\begingroup$ What is your question? $\endgroup$ – John B Mar 8 '16 at 23:38
  • $\begingroup$ @Jonas I suppose I wanted to (i) give such a proof so it may be of help to someone on a search and (ii) inquire why similar proofs are not given elsewhere $\endgroup$ – Anthony Peter Mar 8 '16 at 23:39
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    $\begingroup$ I see. I like your proof. $\endgroup$ – John B Mar 8 '16 at 23:43
  • $\begingroup$ @FaraadArmwood Moreover, I mainly wrote this to show that how it connects to continuity... I figured it may help someone $\endgroup$ – Anthony Peter Mar 8 '16 at 23:45
  • $\begingroup$ @AnthonyPeter sorry, I was thinking of something completely different, ignore that comment. $\endgroup$ – Faraad Armwood Mar 8 '16 at 23:49

Here's somewhat stream-lined take on your argument, avoiding the Cauchy property. Because $\{a_n\}$ converges (to $a$, say), given $\epsilon>0$ there exists $N$ so large that if $n\ge N$ then $a-\epsilon\le a_n\le a+\epsilon$. From the leftmost of these two inequalities, it follows that $\inf_{k\ge n}a_k\ge a-\epsilon$ for all $n\ge N$. Likewise, from the right inequality it follows that $\sup_{k\ge n}a_k\le a+\epsilon$. Putting these together $$ a-\epsilon\le\inf_{k\ge n}a_k\le\sup_{k\ge n} a_k\le a+\epsilon,\qquad\forall n\ge N. $$ Finally, because $\inf_{k\ge n}a_k$ increases to $\liminf_na_n$ and $\sup_{k\ge n} a_k$ decreases to $\limsup_n a_n$ as $n\to\infty$, we get $$ a-\epsilon\le\liminf_n a_n\le\limsup_n a_n\le a+\epsilon, $$ for each $\epsilon>0$.

  • $\begingroup$ Lovely... do you have a source for this? or is this something you just whipped up? I guess I just really want someone learning analysis to know that lim inf and lim sup aren't so mysterious as one might think due to, say, the definition in Baby Rudin, just that they're essentially similar to their continuous analogues... your proof shrouds this a tad, but upon inspection I think one can get the same idea out of it. +1 $\endgroup$ – Anthony Peter Mar 9 '16 at 0:09

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