Sequence Limit (1/n) -> 0

  • #1
The general definition for the limit p of a sequence (pn) is
[itex]\forall \epsilon > 0, \exists N \in \mathbb{N}:n \in \mathbb{N} \geq N \implies d(p_n,p) < \epsilon[/itex]

So, for the experience, I want to derive the limit of the sequence [itex](1/n)_{n \in \mathbb{N}}[/itex]. I would like detailed and harsh criticism on my process. Here are my thoughts:

I want to find an N for each ε such that any natural number greater or equal to N satisfies [itex]-\epsilon < \frac{1}{n} - p < \epsilon[/itex]. This can be rewritten as [itex]p-\epsilon < \frac{1}{n} < p+\epsilon[/itex]. Rewriting the inequality for n instead of 1/n is tricky, but I think this works:

[itex]p-\epsilon < \frac{1}{n} < p+\epsilon \implies
\begin{cases}
\frac{1}{p+\epsilon} < n < \frac{1}{p-\epsilon} & \text{for p ≠ 0 and sufficiently small }\epsilon \\
-\frac{1}{\epsilon} < \frac{1}{\epsilon} < n & \text{for p = 0 and n > 0} \\
n < -\frac{1}{\epsilon} < \frac{1}{\epsilon} & \text{for p = 0 and n < 0} \\
\end{cases}
[/itex]

This works under the assumption that, given p ≠ 0 and ε < |p|, all terms are either positive or negative if p is positive or negative, respectively. Whenever terms of an inequality are of the same sign (positive or negative), their multiplicative inverse also inverts the inequalities. On the contrary, whenever terms of an inequality are of the opposite sign, their multiplicative inverse maintain the same inequality.

Since the goal is to find a N for each ε such that [itex]n \geq N \implies -\epsilon < \frac{1}{n} - p < \epsilon[/itex], n should not be bounded from above lest there be an n > N that does not meet the requirement. [itex]n[/itex] is not bounded above only in the p=0 and n>0 case, and so [itex](\forall ε > 0, \exists N = \left \lceil \frac{1}{ε} \right \rceil: n \geq N \implies -\epsilon < \frac{1}{n} < \epsilon) \implies p=0[/itex].
 

Answers and Replies

  • #2
22,089
3,286
I can't say your proof is bad, it's pretty good actually. But some things can be done better. First a notation issue:

The general definition for the limit p of a sequence (pn) is
[itex]\forall \epsilon > 0, \exists N \in \mathbb{N}:n \in \mathbb{N} \geq N \implies d(p_n,p) < \epsilon[/itex]
Writing ##n\in \mathbb{N}\geq N## makes no sense. You are writing now ##\mathbb{N}\geq N##, which indicates a "set" being larger than a number. This makes no sense. You should write ##n\in \mathbb{N},~n\geq N##.

Anyway, on to another point. What you did is write down the epsilon-delta formula for arbitrary p. You then did some manipulations and then you found what p is. While all of this is pretty good for discovering the limit, it's not how you would actually present the proof.

The way to present the proof is to say "We claim that ##p=0## and thus that ##1/n\rightarrow 0##. And then you should do the entire proof but only with ##p=0##. So all the manipulations you did for arbitrary n, should only be done for p=0.

That said, there is something missing in your proof. Have you ever seen Archimedes axiom (it's not really an axiom)?? You should use this somewhere. The result ##1/n\rightarrow 0## depends on that





So, for the experience, I want to derive the limit of the sequence [itex](1/n)_{n \in \mathbb{N}}[/itex]. I would like detailed and harsh criticism on my process. Here are my thoughts:

I want to find an N for each ε such that any natural number greater or equal to N satisfies [itex]-\epsilon < \frac{1}{n} - p < \epsilon[/itex]. This can be rewritten as [itex]p-\epsilon < \frac{1}{n} < p+\epsilon[/itex]. Rewriting the inequality for n instead of 1/n is tricky, but I think this works:

[itex]p-\epsilon < \frac{1}{n} < p+\epsilon \implies
\begin{cases}
\frac{1}{p+\epsilon} < n < \frac{1}{p-\epsilon} & \text{for p ≠ 0 and sufficiently small }\epsilon \\
-\frac{1}{\epsilon} < \frac{1}{\epsilon} < n & \text{for p = 0 and n > 0} \\
n < -\frac{1}{\epsilon} < \frac{1}{\epsilon} & \text{for p = 0 and n < 0} \\
\end{cases}
[/itex]

This works under the assumption that, given p ≠ 0 and ε < |p|, all terms are either positive or negative if p is positive or negative, respectively. Whenever terms of an inequality are of the same sign (positive or negative), their multiplicative inverse also inverts the inequalities. On the contrary, whenever terms of an inequality are of the opposite sign, their multiplicative inverse maintain the same inequality.

Since the goal is to find a N for each ε such that [itex]n \geq N \implies -\epsilon < \frac{1}{n} - p < \epsilon[/itex], n should not be bounded from above lest there be an n > N that does not meet the requirement. [itex]n[/itex] is not bounded above only in the p=0 and n>0 case, and so [itex](\forall ε > 0, \exists N = \left \lceil \frac{1}{ε} \right \rceil: n \geq N \implies -\epsilon < \frac{1}{n} < \epsilon) \implies p=0[/itex].
 
  • #3
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
412
The general definition for the limit p of a sequence (pn) is
[itex]\forall \epsilon > 0, \exists N \in \mathbb{N}:n \in \mathbb{N} \geq N \implies d(p_n,p) < \epsilon[/itex]
In addition to what micromass said, there should also be a ##\forall## acting on the n. This is one way (but not the only way) to write the definition:
$$\forall \epsilon > 0\ \exists N \in \mathbb{N}\ \forall n \in \mathbb{N} \big(n\geq N \implies d(p_n,p) < \epsilon\big).$$
So, for the experience, I want to derive the limit of the sequence [itex](1/n)_{n \in \mathbb{N}}[/itex]. I would like detailed and harsh criticism on my process.
As micromass said, what you said looks like a description of how to find the proof. Since you're talking about your "process", I'm assuming that what you want to talk about isn't just how to present the proof, but also how to discover it.

I want to find an N for each ε such that any natural number greater or equal to N satisfies [itex]-\epsilon < \frac{1}{n} - p < \epsilon[/itex].
You should probably start by saying "Let ε>0 be arbitrary." What you want to find is a real number p such that there exists a positive integer N such that ##\left|\frac 1 n-p\right|<\varepsilon## for all integers n≥N. So you need to guess what p is first. (I'm saying "guess", because it doesn't matter if you use a rigorous argument or just write down what you think the number should be). In this problem, it's intuitively obvious that p=0. So you don't have to put any more thought into that. Then you use your guess for p to guess an N that gets the job done.

Once you've done that, you verify that your guesses are correct. This verification is the actual proof. The thought process you went through to guess p and N isn't part of the proof. That's why you don't have to use a rigorous argument for that part.

All you need to do to find N is to note that the N you choose must satisfy a very simple inequality. Do you see which one I have in mind?
 

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