Definition of the limit of a sequence

[member 587159]

The definition of a limit of a sequence,
if the limit is finite, is:

lim n >infinity un (un is a sequence) = l

<=>

∀ε> 0, ∃N: n > N => |un - l| < ε

This just means that un for n > N has to be a number for which: l -ε < un < l + ε

Now, I'm wondering, can't we just say:

n > N => |un -l| < kε with k a real number larger than zero?

Thanks in advance

 

[fresh_42]

What for? There is no mathematical difference between $ε$ and $kε$ for a constant $k$. Why do you want to complicate something without need that often is hard to understand to newbies anyway.

 

[micromass]

What for? There is no mathematical difference between $ε$ and $kε$ for a constant $k$. Why do you want to complicate something without need that often is hard to understand to newbies anyway.

I disagree that it is without need. You often see in mathematical derivations that something is $<3\varepsilon$ and that that's good enough. Such proofs are even known as $3\varepsilon$-proofs. I do agree that what's in the OP shouldn't be the standard definition as it already is hard enough. But it should be made much more clearer in analysis books that $<k\varepsilon$ is good enough.

 

[PeroK]

The definition of a limit of a sequence,
if the limit is finite, is:

lim n >infinity un (un is a sequence) = l

<=>

∀ε> 0, ∃N: n > N => |un - l| < ε

This just means that un for n > N has to be a number for which: l -ε < un < l + ε

Now, I'm wondering, can't we just say:

n > N => |un -l| < kε with k a real number larger than zero?

Thanks in advance

When people are learning limits often a lot of complexity comes about for the case where $\epsilon$ is not small. Also, sometimes it complicates things to get the $< \epsilon$ in the definition and it's simpler to get $\le \epsilon$. I always thought a better (equivalent) definition would be.

$\exists \epsilon_0 > 0: \forall \epsilon \in (0, \epsilon_0), \exists N: n > N \Rightarrow |u_n - L| \le \epsilon$

In fact, as I don't have to sit exams, this is the one I use!

You could, if you wanted to, add your $k$:

$\exists k, \ \epsilon_0 > 0: \forall \epsilon \in (0, \epsilon_0), \exists N: n > N \Rightarrow |u_n - L| \le k\epsilon$

 

[fresh_42]

I disagree that it is without need. You often see in mathematical derivations that something is $<3\varepsilon$ and that that's good enough. Such proofs are even known as $3\varepsilon$-proofs. I do agree that what's in the OP shouldn't be the standard definition as it already is hard enough. But it should be made much more clearer in analysis books that $<k\varepsilon$ is good enough.

I learned the notation $\forall \epsilon \exists N(\epsilon)$ or $\forall \epsilon \exists N_\epsilon$ so there has never been a doubt that a the find of $N$ depends on the choice of $\epsilon$ and a constant factor can easily be transformed into $N_\epsilon$.

I find an additional "variable" in the definition would only cause more difficulties, such as to think the small sized $\epsilon$ could be blown up beyond any natural number. Although still correct, it's not helpful. (IMO, but I don't want hell to break lose ...)

 

[fresh_42]

When people are learning limits often a lot of complexity comes about for the case where $\epsilon$ is not small. Also, sometimes it complicates things to get the $< \epsilon$ in the definition and it's simpler to get $\le \epsilon$. I always thought a better (equivalent) definition would be.

I agree, however, wouldn't that simply transform the difficulties to later concepts, because it then makes the difference between an open and a closed neighborhood. I always thought this to be the reason for the strict < . Otherwise one has to explain why all of a sudden we change from closed to open.

 

[micromass]

I agree, however, wouldn't that simply transform the difficulties to later concepts, because it then makes the difference between an open and a closed neighborhood. I always thought this to be the reason for the strict < . Otherwise one has to explain why all of a sudden we change from closed to open.

You're right. But when I first learned analysis at university, one of the very first exercises we did was to show that we could replace all notions of $<$ with $\leq$. I remember that I found it somewhat difficult to grasp, and I didn't really see its importance. In retrospect, I'm happy that we did this.

 

[member 587159]

I disagree that it is without need. You often see in mathematical derivations that something is $<3\varepsilon$ and that that's good enough. Such proofs are even known as $3\varepsilon$-proofs. I do agree that what's in the OP shouldn't be the standard definition as it already is hard enough. But it should be made much more clearer in analysis books that $<k\varepsilon$ is good enough.

Yes, this is exactly why I made this thread. In the following proof, we encounter something similar:

Theorem: If lim_{n-> ∞}u_n = l₁ and lim_{n-> ∞}v_n = l₂

lim_{n-> ∞}(u_n + v_n) = l₁ + l₂

Proof: We can say for all ε>0

∃N: n> N => |u_n-l₁|<ε
∃N': n> N' => |v_n-l₂|<ε

Now, we observe, that:

|u_n + v_n - (l₁ + l₂)| = |(u_n - l₁) + (v_n - l₂)| =< |u_n - l₁| + |v_n - l₂|

So, for n > max{N, N'}, we find:

|u_n + v_n - (l₁ + l₂)| = |(u_n - l₁) + (v_n - l₂)| =< |u_n - l₁| + |v_n - l₂| < ε + ε

=> |u_n + v_n - (l₁ + l₂)| < 2ε

Now, the book does something I'm unsure about:

They say, choose ε' > 0, and let ε = ε'/2. Then we have, for all ε' > 0

n > max {N, N'} => |u_n + v_n - (l₁ + l₂)| < ε' and what we wanted to proof follows.

It makes sense to me that |u_n + v_n - (l₁ + l₂)| < 2ε would be enough to proof, as Micromass already said (that's why I created this thread), but I do not fully understand why we can say that ε = ε'/2? Is this, because ε>0 => ε'/2 > 0?

Probably this is a dumb question, but I'm self studying this to prepare for university mathematics to have at least a bit experience with some more abstract mathematics. I got this analysis book from my brother who studies physics.

 

[fresh_42]

Yes, this is exactly why I made this thread. In the following proof, we encounter something similar:

Theorem: If lim_{x-> ∞}u_n = l₁ and lim_{x-> ∞}v_n = l₂

lim_{x-> ∞}(u_n + v_n) = l₁ + l₂

Proof:

It will get better with time. Wait until you will finish your proofs as soon as some $\epsilon$ show up
and all the work now, like $\epsilon' = 2 \epsilon$, $\epsilon' = \frac{1}{3} \epsilon$ or $\epsilon = \min\{\epsilon_1, \epsilon_2\}$ will be classified under the label: "boring technical issues". Unfortunately until then, the boring stuff has to be learned, too. (see post #7)

 

[member 587159]

It will get better with time. Wait until you will finish your proofs as soon as some $\epsilon$ show up
and all the work now, like $\epsilon' = 2 \epsilon$, $\epsilon' = \frac{1}{3} \epsilon$ or $\epsilon = \min\{\epsilon_1, \epsilon_2\}$ will be classified under the label: "boring technical issues". Unfortunately until then, the boring stuff has to be learned, too. (see post #7)

Well I do not think it's boring at all ;)

 

[PeroK]

Well I do not think it's boring at all ;)

The other aspect of this is the powerful logical thinking required. For example, the negation of the limit. This confounds a lot of people when they first meet it. How do you say that a sequence does not have a limit of $L$? This logical thinking is vital, even if the epsilons and deltas have been replaced by other more abstract concepts.

 

[member 587159]

The other aspect of this is the powerful logical thinking required. For example, the negation of the limit. This confounds a lot of people when they first meet it. How do you say that a sequence does not have a limit of $L$? This logical thinking is vital, even if the epsilons and deltas have been replaced by other more abstract concepts.

I'm not quite sure about this one:

Would this be:

∃ε > 0, ∀N: n>N => |u_n - L | >= ε

or something like..

∀ε > 0, there is no N: n>N => |u_n - L | < ε

EDIT: could someone give a look at post #8?

 

[PeroK]

I'm not quite sure about this one:

Would this be:

∃ε > 0, ∀N: n>N => |u_n - L | >= ε

or something like..

∀ε > 0, there is no N: n>N => |u_n - L | < ε

Neither, I'm afraid. Have a good think!

 

[micromass]

I'm not quite sure about this one:

Would this be:

∃ε > 0, ∀N: n>N => |u_n - L | >= ε

or something like..

∀ε > 0, there is no N: n>N => |u_n - L | < ε

EDIT: could someone give a look at post #8?

Both are incorrect.

 

[Svein]

What for? There is no mathematical difference between $ε$ and $kε$ for a constant $k$. Why do you want to complicate something without need that often is hard to understand to newbies anyway.

Not only newbies. In the University I once had a very old-fashioned professor. When proving a limit, he chose an ε and a δ, and after about half an hour he ended up with ".. less than 2ε". Of course, we were satisfied, but he started the next lecture by choosing a δ half as big as he had done in the previous lecture and did the whole proof over!

On another note, I remember a proof in multidimensional complex analysis which ended up with "... less that 10 000ε, which is small when ε is small."

 

[member 587159]

Neither, I'm afraid. Have a good think!

∃ε > 0, ∀N: n>N ∧ |u_n - L | >= ε

or: ∃ε > 0, ∀N: |u_n - L | >= ε => n=<N

I used neg(a=>b) <=> a Λ neg(b) for the first and contraposition for the second. I had to look up some logic first because this is something we were not taught in high school.

Can someone look at post #8, this is where I am stuck in the proof.

 

[PeroK]

It makes sense to me that |u_n + v_n - (l₁ + l₂)| < 2ε would be enough to proof, as Micromass already said (that's why I created this thread), but I do not fully understand why we can say that ε = ε'/2? Is this, because ε>0 => ε'/2 > 0?

Probably this is a dumb question, but I'm self studying this to prepare for university mathematics to have at least a bit experience with some more abstract mathematics. I got this analysis book from my brother who studies physics.

It's not a dumb question. The crux of the matter is that if you choose any $\epsilon > 0$, then $\epsilon ' = \epsilon/2 > 0$. And, for this $\epsilon '$ you can find $N_1, N_2$ such that:

$n > N_1 \Rightarrow |u_n - l_1| < \epsilon ' = \epsilon/2$

And

$n > N_2 \Rightarrow |v_n - l_2| < \epsilon ' = \epsilon/2$

Now, $n > max\{N_1, N_2 \}$ gives you what you want.

Does that make sense?

Personally, I would do this without introducing $\epsilon '$. I'd just say that because $\epsilon/2 > 0$ we can find an $N_1$ such that:

$n > N_1 \Rightarrow |u_n - l_1| < \epsilon/2$ etc.

PS I can see the book did things a bit differently. I definitely don't like using $N'$ for the $l_2$ limit. Using $N_1, N_2$ seems much more logical. Especially as $N'$ has nothing to do with $\epsilon '$

 

[member 587159]

It's not a dumb question. The crux of the matter is that if you choose any $\epsilon > 0$, then $\epsilon ' = \epsilon/2 > 0$. And, for this $\epsilon '$ you can find $N_1, N_2$ such that:

$n > N_1 \Rightarrow |u_n - l_1| < \epsilon ' = \epsilon/2$

And

$n > N_2 \Rightarrow |v_n - l_2| < \epsilon ' = \epsilon/2$

Now, $n > max\{N_1, N_2 \}$ gives you what you want.

Does that make sense?

Personally, I would do this without introducing $\epsilon '$. I'd just say that because $\epsilon/2 > 0$ we can find an $N_1$ such that:

$n > N_1 \Rightarrow |u_n - l_1| < \epsilon/2$ etc.

PS I can see the book did things a bit differently. I definitely don't like using $N'$ for the $l_2$ limit. Using $N_1, N_2$ seems much more logical. Especially as $N'$ has nothing to do with $\epsilon '$

In fact, the book used N₁ and N₂ but I wrote down the proof without the book. You are right that this is more logical. I believe your explanation helped me further. Thank you for that! I will think about it some more. In the meantime, what do you think about the negation of the limit?

 

[PeroK]

In fact, the book used N₁ and N₂ but I wrote down the proof without the book. You are right that this is more logical. I believe your explanation helped me further. Thank you for that! I will think about it some more. In the meantime, what do you think about the negation of the limit?

For the negation, think about the sequence:

$1, 1/2, 1, 1/4, 1, 1/8, \dots$

And why it does not converge to either $0$ or $1$ (or anything else!).

 

[member 587159]

For the negation, think about the sequence:

$1, 1/2, 1, 1/4, 1, 1/8, \dots$

And why it does not converge to either $0$ or $1$ (or anything else!).

Is what I wrote in post #16 wrong?

 

[PeroK]

Is what I wrote in post #16 wrong?

The first one is close, although I'm not quite sure what it means.

 

[member 587159]

The first one is close, although I'm not quite sure what it means.

∃ε > 0, ∀N: n>N ∧ |un - L | >= ε

∧ is the logical symbol for 'and'
>= means 'greater than or equal to epsilon'

 

[PeroK]

∃ε > 0, ∀N: n>N ∧ |un - L | >= ε

∧ is the logical symbol for 'and'
>= means 'greater than or equal to epsilon'

Read it out to me, then!

"There exists ..."

 

[fresh_42]

It might help to start with convergence defined by
$\forall \epsilon > 0 \;\; \exists N_\epsilon \in \mathbb{N} \;\; \forall n > N_\epsilon \;\; (|u_n - L| < \epsilon)$

 

[fresh_42]

On another note, I remember a proof in multidimensional complex analysis which ended up with "... less that 10 000ε, which is small when ε is small."

A side note: This reminds me on people who write everything constant as $\mathcal{O}(1)$.

 

[member 587159]

Read it out to me, then!

"There exists ..."

The limit of a sequence is not equal to L
if and only if
there exists an epsilon larger than zero and we can say for all N, that: n is larger than N and |u_n - L| is larger or equal to epsilon.

 

[fresh_42]

The limit of a sequence is not equal to L
if and only if
there exists an epsilon larger than zero and when we can say for all N, that: n > N and |u_n - L| is smaller or equal to epsilon.

You are repeating what you already said. There is a quantifier missing. It might help to negate what I've written in #24.

 

[member 587159]

You are repeating what you already said. There is a quantifier missing. It might help to negate what I've written in #24.

∃ε > 0, ∀N, ∃n>N: |un - L | >= ε

I believe this is the negation.

 

[PeroK]

The limit of a sequence is not equal to L
if and only if
there exists an epsilon larger than zero and when we can say for all N, that: n > N and |u_n - L| is smaller or equal to epsilon.

I still don't know what that means. As an antidote to this symbolic stuff:

A sequence does not converge (to L) if there is a finite distance from L such that no matter how far you go along the sequence, at some later point it will be further than this distance from L.

I would try to think about that geometrically as well: L as a line, $\epsilon$ as the width of a tunnel either side of L and the sequence (at least at some points) continually going outside the tunnel.

Whereas, if it converges to L, then no matter how narrow you make the tunnel, the sequence eventually goes into the tunnel and stays there!

 

[fresh_42]

∃ε > 0, ∀N, ∃n>N: |un - L | >= ε

I believe this is the negation.

You should (in order to practice) apply this to Perok's sequence and show why it does not converge.

 

[Svein]

Definitions written in symbols:

Given a sequence x_n, we say that x is a cluster point for x_n if $(\forall \epsilon >0)(\forall N)(\exists n>N)(\vert x-x_{n} \vert <\epsilon)$.

Given a sequence x_n, we say that x_n converges to x if $(\forall \epsilon >0)(\exists N)(\forall n>N)(\vert x-x_{n} \vert <\epsilon)$.