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Homework Help: Rigorous proof of limits of sequences

  1. Jan 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Definition: Let an be a sequence of real numbers. Then an->a iff
    for all ε>0, there exists N such that nN => |an - a|<ε.

    Let an=(n2+1)/(n2-9).
    PROVE that an->1 as n->∞.

    Proof:
    Assume n≥4. Then | 1-an | = 10/(n2-9).
    10/(n2-9) < 10/n provided n2 - 9 > n, i.e. n2 - n > 9.
    Now n2-n=n(n-1) ≥n for n≥2
    and so n2-n > 9 for n>9

    10/(n2-9) < 10/n < ε provided n>9 AND n>10/ε. Thus take N=max{9,10/ε}.
    ======================

    1) In the first place, WHY can we assume n≥4 at the beginning of the proof?

    2) Secondly, I don't think the choice of N above works.
    N=max{9,10/ε} => N≥9 and N≥10/ε
    In the definition of limit, we have nN, so this implies n≥N≥9 and n≥N≥10/ε, so n≥9 and n≥10/ε, but this does NOT imply n>9 AND n>10/ε.
    But the inequalities 10/(n2-9) < 10/n < ε is only justified provided n>9 AND n>10/ε.

    So is this proof wrong? In the definition of limit, can we use n>N instead of nN?

    2. Relevant equations
    N/A

    3. The attempt at a solution
    N/A

    Thanks for any help! :)
     
  2. jcsd
  3. Jan 8, 2010 #2
    I looked over the proof and each step seemed justified. As for your concerns,

    1) Well we can of course require n to be bigger than any given number, since we only care about what happens when n is big. This shouldn't startle you, since we require n to be bigger than 9 in the end anyways. Presumably, the author of the proof wanted to ensure n^2 - 9 > 0. Thus there is really no harm in requiring n to be bigger than 4.

    2)A minor note: N = max{9, 10/e} does not mean N [itex]\geq[/itex] 9 , N [itex]\geq[/itex]10/e. You have the inequalities reversed. Remember N is the bigger of the two numbers. Regardless I still think the argument is valid since n[itex]\geq[/itex]N is n > N OR n = N, and we can simply take n > N. But let's face it, this is a very tiny point. Many authors stipulate that n > N in the definition. The important thing you should take away from this is that n[itex]\geq[/itex]N is more "lenient" since n could be just > N (it's easy to think that it should be the opposite, as I think you have done here).
     
  4. Jan 8, 2010 #3
    First off, I have one question: if you start with n≥4, so why do you end up with n≥9 which makes the assumption contradictory?

    Actually the choice of n≥4 is flawless; you can see that for instance n=3 the denominator in a_n is zero, leading to an absurdity. So let alone numbers less than 3 or equal to it and start with n≥4. We CANNOT make a choice of the form n≥1 but n=/=3. One only needs to realize the roots of denominator then go ahead with numbers greater than the positive root (if it is an integer).

    Second stuff to be noticed is that the statement "10/(n2-9) < 10/n provided n^2 - 9 > n, i.e. n2 - n > 9" is wrong. It is easy to check that 10/(n^2-9) < 10/n is compatible with n≥4. You got to solve the inequality than to only behold the face and judge.

    10/(n2-9) < 10/n < ε provided n>9 AND n>10/ε. Thus take N=max{9,10/ε}.

    During the proof, you make a lot of paradoxical assumptions/reasonings: You must take into account the prior bounds for n to not go beyond 'em afterwards. Now it is crystal clear that N=max{4,10/ε} is N=4.

    But note that for N=4 the sequence gives a_N= 16/7[tex]\approx[/tex]2.28, thus 1.28<ε. You don't be worried about this. It is only because the choice of n≥4 does not sound so extraordinary. If it was n≥5, then we woud have 0.625<ε as if ε is upper-bounded by 1, this would be much realistic.
    [/QUOTE]
     
  5. Jan 8, 2010 #4

    HallsofIvy

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    What? How does "[itex]n\ge 9[/itex]" contradict [itex]n\ge 4[/itex]? Any number that is [itex]\ge 9[/itex] is [itex]\ge 4[/itex].

    That is not only not "crystal clear", it is not necessarily true. If [itex]\epsilon= .0001[/itex], then [itex]10/\epsilon= 1000000[/itex] and [itex]N= max(4, 10/\epsilon)= 100000[/itex].

    Nonsense- the whole point is that [itex]\epsilon[/itex] is given and that [itex]\epsilon[/itex] can be any number greater than 0. You cannot assert that [itex]\epsilon[/itex] is any particular number.

     
    Last edited by a moderator: Jan 8, 2010
  6. Jan 8, 2010 #5
    Yes, but the inverse does not hold. We first assume [itex]\ge 4[/itex] then we get [itex]\ge 9[/itex]. So what about n=7!? Considering [itex]\ge 9[/itex] from the beginning won't get to any such misleading. One needs to go for 'good' choices of n to not probably hit a snag ahead.

    I take it as a typo that you give us two different answers 100000 and 1000000. Pardon me, but if [itex]\epsilon= .0001[/itex], then what about [itex]10/(n^2-9) < 10/n < \epsilon[/itex]===>[itex]10/(n^2-9) < 10/9< 0.0001 [/itex]?

    I don't remember I said ε is NOT given. What I just claimed was that the OP should be aware of his choice of n. Period.

    AB
     
    Last edited: Jan 8, 2010
  7. Jan 8, 2010 #6

    HallsofIvy

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    Then you completely misunderstood the proof. First he pointed out that n must be greater than or equal to 4, just in order to be defined. Then he said that, in order that he be able to use a certain inequality, n must also be larger than or equal to 9.



     
  8. Jan 8, 2010 #7
    \geq
    Ugh... I thought he meant to use [tex]n\geq 4[/tex] as a first guess for N! My bad.
     
    Last edited: Jan 8, 2010
  9. Jan 8, 2010 #8
    But I think N=max{9,10/ε} => N≥9 and N≥10/ε is absolutely correct.

    You said the inequalities should be reversed. So "N=max{9,10/ε} => N<9 and N<10/ε". But this is incorrect......
     
  10. Jan 8, 2010 #9
    Thanks for the inputs. But I still don't understand the two concerns that I posted in my first post and I would appreciate if someone can explain these.

    1) In the first place, WHY can we assume n≥4 at the beginning of the proof? Why are we even allowed to do this? Can we assume n≥10 at the beginning?

    2) Secondly, I don't think the choice of N above works. (please pay attention to whehter the inequalities are "strict" inequalities or "weak" inequalities)

    N=max{9,10/ε} => N≥9 and N≥10/ε
    In the definition of limit, we have nN, so this implies n≥N≥9 and n≥N≥10/ε, so n≥9 and n≥10/ε, but this does NOT imply n>9 AND n>10/ε. (n>9 implies n≥9 is correct, but n≥9 => n>9 is wrong)
    The inequalities 10/(n2-9) < 10/n < ε is only justified provided n>9 AND n>10/ε, but if we choose N=max{9,10/ε}, this only implies n≥9 and n≥10/ε.

    So is this proof wrong? In the definition of limit, can we use n>N instead of nN? Would this be an exactly equivalent definition?

    Thank you!
     
  11. Jan 8, 2010 #10
    Actually he is right about his claim of N being the bigger of the two numbers, so here one gets N=10/ε. This result and n≥9 and that ε>10/(n^2-9) gives N < n^2-9 or N<72. But I still think that your choice of N is not 'good'...
     
  12. Jan 8, 2010 #11
    Sounds like you are making my mistake, too!

    1- Here n≥4 is not an initial guess for N. It just shows that the expression in the absolute value funstion is positive for all required n's, so you can easily write it without |...|. As you may have noticed, for n<3 a_n itself gives us negative values so let alone n's smaller than 4. And we are JUST allowed to take n≥4 as a guess if it does not make ε get larger enough to say confidently that the sequence does not converge. And Yes you are allowed to even take n≥100 in the first place but it may exceed the minimum 'safe' value for N that we usually do look for it.

    2- I agree, because that gives ε>10/72 which is not a 'safe', but yet acceptable, value for epsilons that are used to be much more smaller than that. Let's say if you had started with n≥20, you would have been in a more 'safer' zone... :)

    I think these can help you!

    AB
     
  13. Jan 8, 2010 #12

    vela

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    You can assume n is greater than any fixed number M because the first M terms of the sequence don't tell you anything about whether the sequence is going to converge. You only care about what happens with the sequence when n goes to infinity. In other words, if you had a new sequence which was just the original sequence without the first M terms, the new sequence converging implies the original sequence converges.

    You are correct the implications don't follow, but you could easily fix the proof by using N=max{9,10/ε}+1 instead.

    You can use n>N in the definition of the limit. I think this is actually how it's usually done to avoid the sort of technicality you've noticed.
     
  14. Jan 8, 2010 #13

    vela

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    That's not how it works. This isn't like doing an experiment and figuring out small the error bars have to be to be acceptable. [tex]\epsilon[/tex] is a given. It's some arbitrary positive number. It could be [tex]10^{100}[/tex], [tex]10^{-100}[/tex], or any other positive value. The proof is just showing that if you go far enough out in the sequence, then the terms are bounded between [tex]L-\epsilon[/tex] and [tex]L+\epsilon[/tex], where [tex]L[/tex] is the limit of the sequence. How far into the sequence you have to go depends on [tex]\epsilon[/tex], but the point is that you can do this for any value of [tex]\epsilon[/tex], no matter how small it is.
     
  15. Jan 8, 2010 #14
    1) OK, now I get the point. So at any point of the proof, we can assume n is greater than any fixed number M to obtain estimates, and we just have to make sure that at the end we must choose N to be greater than (or equal to) M.

    2) Thanks, I think you have answered my questions.
    So I think the two definitions below are equivalent and are both acceptable.
    "For all ε>0, there exists N such that nN => |an - a|<ε."
    "For all ε>0, there exists N such that n>N => |an - a|<ε."

    Another question: in this definition, does N have to be an integer? Why or why not? If the answer is yes, then I think the proof above will have some problem, becuase 10/ε is not necessarily an integer?

    Thanks! :)
     
  16. Jan 8, 2010 #15

    vela

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    No, N doesn't have to be an integer. It's just some real number, for a given [tex]\epsilon[/tex], that integer n must exceed for the limit implication to hold.
     
  17. Jan 8, 2010 #16
    But in MANY textbooks, they require N to be an integer.
    Wikipedia does the same. (http://en.wikipedia.org/wiki/Limit_of_a_sequence#Formal_definition)

    What is the reason of requiring N to be an integer? I don't think it is absolutely necessary...but since so many textbooks are doing it, I think there must be a reason?
     
    Last edited: Jan 8, 2010
  18. Jan 8, 2010 #17

    vela

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    That's interesting. I don't know of a reason offhand. Maybe it's some technical point about comparing n to N relying on the ordering of the natural numbers.

    You could modify your proof to say "Choose [tex]N\in\mathbb{N}[/tex] such that [tex]N > max\{9,10/\epsilon\}[/tex]." That would also take care of your concern about the [tex]\ge[/tex] vs. [tex]>[/tex].
     
  19. Jan 9, 2010 #18
    It is not like that! Keep in mind that [tex]\epsilon[/tex] measures the error between the numbers a_n and the limit L, while the integer N measures how fast the sequence gets closer to the limit L (as you said). So what I mean is that while [tex]\epsilon[/tex] is a given, you can't choose [tex]\epsilon=10^{100}[/tex] ever. That is a terrible, no let's say, a very very terrible choice of [tex]\epsilon[/tex], that won't work out. In the definition of limit, we say [tex][f(x+\epsilon)]-f(x)]/\epsilon[/tex] where [tex]\epsilon[/tex] goes to zero to only assure us that the derivative of f(x) exists at x. Most authors don't emphasize this in their definition of limit of a sequence, but others, while doing so in the definition itself, prefer to notify readers as a tip or something of this critical point that [tex]\epsilon[/tex] is a very small number.

    It is just because your list of numbers for N would be so narrowed down so you can feel really happy that you won't get into the trouble of what you should choose from the scary set of real numbers!:rofl:
     
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