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Limit proof that 1=0 - Where is the error?

  1. Dec 28, 2013 #1
    Limit proof that 1=0 -- Where is the error?

    [itex]\lim_{n\to\infty } 1=\lim_{n\to\infty }\frac{n}{n}=\lim_{n\to\infty }\frac{\overbrace{1+1+\ldots+1}^{n \text{ times}}}{n}=\lim_{n\to\infty }\frac{1}{n} + \lim_{n\to\infty }\frac{1}{n} + \ldots=0[/itex]

    I have a feeling it is in the step where you split the limit of a sum into the sum of the limit of the terms, but I do not have any exact explanation as to why. Can anyone help?
  2. jcsd
  3. Dec 28, 2013 #2

    D H

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    The limit of a sum is the sum of the limits on the individual terms if the number of terms is finite and if each of the terms has a limit. This is the algebraic limit theorem. Your splitting n into 1+1+1…+1 violates condition #1. What you did is invalid.
  4. Dec 28, 2013 #3
    Yeah, it's a very good one though. The error is simply because we are taking the limit as n approaches infinity, making this actually an infinite sum. With any other limit, doing these steps will result in no problems.
  5. Dec 28, 2013 #4
    What is the mathematical reason as to why you can't split the limit when there is an infinite sum?
  6. Dec 28, 2013 #5
    I don't think the infinite sum is the issue here. There are theorems (like monotone convergence theorem) which thell you that you can split up some infinite sums and limits.

    Furthermore, you never actually have an infinite sum in the OP.

    What you did is:

    [tex]\lim_{n\rightarrow +\infty} \sum_{k=1}^n \frac{1}{n} = \sum_{k=1}^n \lim_{n\rightarrow +\infty} \frac{1}{n}[/tex]

    The problem is that the sum in the LHS depends on ##n##, which is the variable in the limit. You can't just pull something that depends on the variable of the limit outside of the limit. It's the same thing as saying

    [tex]\lim_{x\rightarrow 0} x = x \lim_{x\rightarrow 0} 1 = x[/tex]
  7. Dec 28, 2013 #6
    Well, that's a convincing explanation, but you can easily see that all of the steps will cause no problems for any finite limit. That seems to suggest to me that the problem is related to an infinite sum.
  8. Dec 28, 2013 #7
    OK, but there is no actual infinite sum in the OP. However, I do see what you mean. If you change the OP to:

    [tex]\lim_{n\rightarrow +\infty} \sum_{k=1}^{+\infty} \frac{1}{n} = \sum_{k=1}^{+\infty}\lim_{n\rightarrow +\infty} \frac{1}{n}[/tex]

    then this is also wrong and then the problem is exactly the exchange property for infinite sums.

    Some infinite sums can be exchanged though: http://en.wikipedia.org/wiki/Monotone_convergence_theorem#Convergence_of_a_monotone_series But the above does not mean the criteria of the wiki post.
  9. Dec 28, 2013 #8
    How is there not an infinite sum?

    [itex]\lim_{n\to\infty }\Sigma ^{n}_{k=1}f(x)[/itex]


    [itex]\Sigma ^{\infty}_{k=1}f(x)[/itex]

    are the same thing. It's written the top way in the OP, but that's still an infinite sum.
  10. Dec 28, 2013 #9
    They are only the same thing if ##f(x)## does not depend on ##n##. In this case, "##f(x) = 1/n##" depends on ##n##, which is also the upper index of the summation. So we have

    [tex]\lim_{n\rightarrow +\infty}\sum_{k=1}^{n} f(n)[/tex]

    which is clearly not the same thing as

    [tex]\sum_{k=1}^{+\infty} f(n)[/tex]
  11. Dec 28, 2013 #10
    If this is the case then why does it hold for any finite limit?
  12. Dec 28, 2013 #11
    Ok, so those two expressions are not the same, but the point is that the first one is still an infinite sum, just not that one. How else can we interpret this expression?

    If it is not an infinite sum, it must be a finite sum. Can you tell me how many terms are in it, then?
  13. Dec 28, 2013 #12
    Because there is no dependance on ##n## with finite sums. If there is a dependence of ##n##, then the result doesn't hold.

    So something like

    [tex]\lim_{n\rightarrow +\infty} \sum_{k=1}^{10} \frac{1}{n} = \sum_{k=1}^{10} \lim_{n\rightarrow +\infty} \frac{1}{n}[/tex]

    because the summation is finite and doesn't depend on ##n##.

    If you want to know why the equality holds in this case, then you can prove it using analysis techniques and epsilonics. It's a good exercise to see what goes wrong in the infinite case and in the dependence on ##n## case (here the RHS isn't even well-defined).
  14. Dec 28, 2013 #13
    Well, I would argue it's not a sum to begin with. Rather, it's the limit of a sum. So it's the limit of a sum with ##n## terms and where the limit depends on ##n##.
  15. Dec 28, 2013 #14
    Why is this incorrect? Don't they both evaluate to zero? Also, what are the conditions for which you can exchange the infinite sum and the limit?
  16. Dec 28, 2013 #15
    I don't understand how the limit of a sum with ##n## terms as ##n## approaches infinity doesn't qualify as an infinite sum, regardless of how the summands change with the limit value.
  17. Dec 28, 2013 #16
    Because an infinite sum or series is by definition something of the form

    [tex]\lim_{n\rightarrow +\infty} \sum_{k=1}^n a_k[/tex]

    and not of the form

    [tex]\lim_{n\rightarrow +\infty} \sum_{k=1}^n a_{n,k}[/tex]

    I will gladly admit that I'm wrong if you can find a reference where the above is called an infinite sum or series.
  18. Dec 28, 2013 #17
    Why can you not split the limit in the case of an infinite sum?
  19. Dec 28, 2013 #18
    Not sure what will satisfy you. You can not split the limit of an infinite sum because you can find counterexamples to it.
    Furthermore, the proof of the finite sum-case fails with infinite sums.
  20. Dec 29, 2013 #19
    [strike]But what I'm saying is there is no meaningful distinction between what you and DH are saying. You can choose not to call this an infinite sum if you'd like, but I think we can agree that as the limit approaches infinity, the number of summands approaches infinity.

    DH says it's because there are infinite terms as the limit approaches infinity, and we cannot break a limit of sums into a sum of limits if the number of terms is infinite.

    You say it's because we cannot switch the sum and the limit operation order because the sum depends on ##n## which is also used in the function, but then this explanation admits that the sum depends on ##n## because the sum is not finite.

    It sounds like you and DH are pointing out the same problem with the OP's equation with the caveat that "I don't want to call this an infinite sum." In both cases the underlying problem is that there is an 'infinite sum' appearing, you've offered elaboration on why that causes additional problems in this example besides the simple fact that "we can't break the infinite sum" but it's still because there is an 'infinite sum.'

    At the same time, I do understand that we can have a case where the equality fails, your explanation applies, but the sum is not infinite (by any stretch.) But in this case, your explanation applies because the sum is 'infinite.'[/strike]

    I changed my mind, I don't think that the limit/sum is infinite matters at all, and I think your explanation is right and does not depend on anything.

    It is clear that:

    [tex]\lim_{n\rightarrow m} \sum_{k=1}^n \frac{1}{n}[/tex]


    [tex] \sum_{k=1}^n \lim_{n\rightarrow m} \frac{1}{n}[/tex]

    are totally different things, and it has nothing to do with infinity and everything to do with the fact that n is used in both the sum and limit.

    Last edited: Dec 29, 2013
  21. Dec 29, 2013 #20


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    Nope. The lefthand side is undefined because the sum diverges for all ##n##. The way it's written, you have to take the sum first for finite ##n## and then take the limit of that result as ##n \to \infty##.
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