Using sequences to prove a limit doesn't exist?

Question

I am struggling to understand this relationship between sequences/subsequences/limits.

I posted a question here, read the Wiki here, saw this other MSE answer here, and even read this page here. Despite all this I am simply not even getting this concept even a little bit. It's all so dense and abstract and I simply can't make heads or tails of it.

I don't understand how this concept is used, how it works, what it illustrates, what it may or may not prove, etc.

Is this technique mainly to check if a limit does or does not converge to something? Do we need to know what that something is? Or is it a general check for non-convergence? How does it work? Are there any easily understood examples? I feel like I need a really dumbed-down explanation for this because I am simply not getting any of it.

Answer 1

Let consider for simplicity at first the case of sequences with finite limit L.

By the definition of limit, we say that

$$n\to +\infty \quad a_n\to L$$

when

$$\forall \epsilon>0 \quad \exists \bar n \quad\text{such that}\quad \forall n>\bar n \quad |a_n-L|<\epsilon$$

Now the key point is that if the limit exists, for a fixed $\epsilon >0$, it does not matter how we choose $n\to+\infty$ for $n>\bar n$, the condition $|a_n-L|<\epsilon$ will always be satisfied in the sense that $a_n$ will approach the limit L.

Hence when limit exists you cannot find subsequences of $a_n$ with a different limit.

Therefore you can use this fact to prove that a sequence has not limit by finding at least 2 subsequences with different limit. If you can do it, it follows that the given sequence has not limit.

The same argument can be easily generalized to the infinite limit $L=\pm\infty$ of sequence and to the limit of functions.

EDIT

Question: How do we prove that if there exist two subsequences of $a_n$ with different limits, then $|a_n−L|<\varepsilon$ won't always be true?

Suppose you find two subsequences $a_{f(n)}$ and $a_{g(n)}$ such that

$$a_{f(n)}\to L_f \quad a_{g(n)}\to L_g$$

then if we set

$$\varepsilon=\frac{|L_f-L_g|}{3}$$

by definition of limit for the two subsequences we can find

$$\bar n_f\quad\text{such that}\quad \forall n>\bar n_f \quad |a_{n_f}-L_f|<\epsilon$$

$$\bar n_g\quad\text{such that}\quad \forall n>\bar n_g \quad |a_{n_g}-L_g|<\epsilon$$

thus exists

$$\bar n=\max\{ n_f, n_g\}\quad\text{such that}\quad \forall n>\bar n \quad \implies |a_{n_f}-L_f|<\epsilon \quad \land \quad|a_{n_g}-L_g|<\epsilon $$

with

$$(L_f-\varepsilon,L_f+\varepsilon)\cap(L_g-\varepsilon,L_g+\varepsilon)=\emptyset$$

which is in not consistent with the definition of limit for $a_n$.

Answer 2

My understanding of sequences is very concrete: I imagine sequences in $\mathbb{R}^2$ viewed as the local surface of the Earth - Europe, say.

If the sequence $(a_1, a_2, a_3, \dots, a_n, \dots)$ has a limit $L$ as $n \to \infty$, then "eventually the terms of $a_n$ get and stay very close to $L$" for whatever "very close" you wish to pick. Or, concretely, if you hop from place to place in Europe, we say your journey has a "limit" if there's a point from which eventually your hops never stray far away, however small you pick "far away" to mean.

So if the sequence converges to $L$, then so does any infinite subsequence. Indeed, the terms of the main sequence get and stay very close to $L$; and every term of the subsequence is a term of the main sequence; so every term of the subsequence must get and stay very close to $L$. Concretely, if your journey hops from place to place, and my journey lags behind yours, sometimes missing out some of your stops, I'll still eventually come near to your limit point. Because I'm only hopping in your footsteps, and because eventually you never get far away from your limit point, I also eventually won't get far away from your limit point.

This gives us a good way to check if a sequence does not converge: if we can find a subsequence which converges to $A$, and another subsequence which converges to $B$, and $A \not = B$, then the parent sequence can't converge. (If it did, then all its terms would eventually get and stay very close to $A$, and also would get and stay very close to $B$, which is a problem because $A$ and $B$ are some distance apart from each other. I can't be forever hopping around very near to both London and Paris at once.)

Note that that test for non-convergence ("find two different limits which subsequences converge to") doesn't require you to guess some limit to which the main sequence converges. If I can show that someone can arrange to stay always near Paris by choosing some of your resting places, and I can also show that someone can arrange to stay always near London by choosing some of your resting places, then without any further knowledge I can guarantee that your journey didn't converge to a limit.

Answer 3

Okay. The classic example is the flea hopping across a desk. Each hop takes him half way across the remaining distance. He never gets to the end.

So imagine a guy keeping a list of all the fleas hops.

$h_1 = \frac 12$

$h_2 = \frac 34$

$h_3 = \frac 78$

.....

$h_k = \frac {2^k -1}{2^k}$

....

This never ends. The guy keeping record of it knows it will never end.

But the guy keeps records anyway and says. "The flea is definitely approaching the value $1$. That is his limit." And the skeptic in the audience says "But he'll never get to $1$ so it isn't $1$." And the guy keeping records says. "I never said he would get to $1$. I said he will approach $1$. His limit is $1$".

Okay, Now the guy keeping records had a lazy, assistant. The assistant kept tracks of position but not all of them. The assistants records went like this.

$a_1 = \frac 12$

$a_2 = \frac 78$ (he skipped $\frac 34$)

$a_3 = \frac {15}{16}$

$a_4 = \frac {1023}{1024}$ (he skipped several)

.....

etc.

Each record on the assistant list is a record on the master list.

So this is a subsequence.

But even though the assistant skips hops, the assistant does keep recording forever. And even though the assistant doesn't have all the hops, it's still clear the flea is approaching and never gets to $1$.

(How could it be otherwise? It's a subset of the same values.)

So the assistant says "The limit is $1$".

Now let's suppose there are 1 recorder, an obsessive recorder with an agenda, and a lazy sub-assistant to the assistant with the agenda.

And suppose the flea is drunk.

The flea hops all over the place.

$h_1 = \frac 7{59}$

$h_2 = \frac 1{\sqrt{2}}$

$h_3 = 37.5$

$h_4 = .9999$

$h_5 = .5782$

The main guy says "There's no limit. The flea is going everywhere."

The assistant with the agenda decides to only mark the hops that are get closer to $1$.

She records the $\frac 7{59}$. And she records the $\frac 1{\sqrt 2}$ because that is closer to $1$. She ignores the $37.5$ because that is not closer to $1$. She records the $.9999$, then she ignores the next few dozen hops but at $h_39$ she records $.99995782$. Now, somehow, (it didn't have to happen but it did) the flea kept hopping and every not and then it would hop to a point closer to $1$ then it ever had before and the assistant with an agenda recorded it.

"In my sequence, which is a subsequence of yours," she says, "The limit is $1$".

The main assistant says "But it's not in general."

And she says "No, you sequence is $\frac 7{59},\frac 1{\sqrt{2}},37.5,.9999,.5782....$ and does not converge, but my sequence is $\frac 7{59},\frac 1{\sqrt{2}}, .9999, .99995782, .9999999487, etc.$ is a subset of yours, but mine converges to $1$. It's possible for a sequence that doesn't converge to have a subsequence that does."

"Yes," says the main guy, "I guess that is true. So what about your subassistant who took a subsequence of yours. What did he get?"

"Oh, I don't care" said the assistant with an agenda, "Since mine got closer and closer to $1$ and he took his values sporadically from mine, he also has to get closer and closer to $1$, right".

"I guess so" says the main guy, "but you didn't get the same results as me and you took yours from mine...."

"That's because you had no results. If you did I would have had to have come to the same limit, but as you had no limit, I could pick and choose. In fact it might have been possible for someone else to get a subsequence that had a different limit than mine."

"Hmm," says the main guy, "I notice that by some remarkable bit of unlikely probability that the flean never landed an any distance between $23.7$ and $31.5$. So if you wanted to get a subsequence converging to, say, $28$, could you?"

"No," admitted the agendad assistant, "a sequence that doesn't converge can have a subsequence that does converge, but it doesn't have to. But any sequence that does converge to a limit must have all its subsequence converge to the same limit."

"Hmm," said the main guy, "there's a moral in here somewhere."

"Probably not," said the assitant, "It's probably just math. Let's go get lunch."

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Slightly more formal but still (hopefully casual).

$a_k \to M$ means.... there'll come a point where all the $a_k$ are very close to $M$

But for any subsequence, all the terms are in the sequence so the terms of the subsequence must also all get very close to $M$.

Well, the terms of a sequence can't get very close to two different things because any two different things are a finite distance apart and "very" close can get much closer than that.

Let $\{a_n\} \to L$. Let $\{b_n\}$ be a subsequence of $\{a_n\}$. Let $M \ne L$. Let $|M-L|=d > 0$

There'll come a point where all the $a_k$ past that point are closer than $\frac d{10000}$ to $L$. That includes all the $b_n$ as well. So all $b_n$ are closer than $\frac d{10000}$ to $L$. So they are further than $\frac {9999 d}{10000}$ from $M$.

So it is impossible for $b_n \to M$.

And as $a_n$ gets "infinitely close to" $L$, then all the $b_n$ must also get infinitely close to $L$. So $b_n \to L$.