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What is the sum of infinite number of zeros?

  1. May 30, 2014 #1
    dear all, lets consider a length L and divide it into N number of small segments uniformly. Then the length of every segment should be L/N.
    then we add these segments up, which is
    L=[itex]\sum[/itex][itex]\frac{L}{N}[/itex]
    then we take the limit N→[itex]\infty[/itex] at both sides, this means
    [itex]lim_{N\rightarrow\infty}[/itex]L=[itex]lim_{N\rightarrow\infty}[/itex][itex]\sum[/itex][itex]\frac{L}{N}[/itex]

    the left hand is still L, and right hand will become, at the limit, [itex]\underbrace{0+0+0+...}_{infinite}[/itex]

    therefore, the sum of infinite number of zeros will lead to a finite number ? This is my question. I hope someone could give some idea. Thank you !
     
  2. jcsd
  3. May 30, 2014 #2

    phinds

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    Your problem is that you are trying to use infinity as a number (so that you can get n/infinity = 0) and it isn't working out for you, which really is just what you should expect. In other words, you're mixing apples and oranges and going bananas. :smile:
     
  4. May 30, 2014 #3
    So I can not substitute the infinity into the L/N, right ?
     
  5. May 31, 2014 #4

    D H

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    No, you can't. Keep in mind what ##\sum_{n=1}^{\infty} a_n## means. It is the limit (if it exists) of the partial sums as the number of terms gets ever larger. Mathematically, ##\sum_{n=1}^{\infty} a_n \equiv \lim_{N\to\infty} S_N## where ##S_N =\sum_{n=1}^N a_n##.

    If all ##a_n## are zero then each partial sum is zero, so the limit is zero.
     
  6. May 31, 2014 #5
    ok, I know that I can not substitute the [itex]\infty[/itex] into L/N when N->∞, but I still feel it is natural to think the sum [itex]\sum[/itex][itex]^{n=1}_{N}[/itex][itex]\frac{L}{N}[/itex] as a addition of infinite zeros, when N-> ∞.
    In fact I encounter the similar situation in the calculus. The integration of f(x) on the domain (a, b) is defined as
    [itex]\int[/itex][itex]^{a}_{b}[/itex] f(x)dx=lim[itex]_{N->\infty}[/itex][itex]\sum[/itex][itex]^{N}_{i=1}[/itex] f(x[itex]_{i}[/itex])dx[itex]_{i}[/itex], (1)
    where N is the divisions of the domain (a, b).
    The same situation as before, because when N->∞, dx[itex]_{i}[/itex]-> 0, I think it is very natural to think the right hand of equ. (1) as a sum of infinite zeros, namely, [itex]\underbrace{0+0+0+...}_{infinite}[/itex]
     
  7. May 31, 2014 #6

    HallsofIvy

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    You are welcome to think of it that way, if it helps. But don't expect the usual laws of arithmetic to help.
     
  8. May 31, 2014 #7
    hello, friends, thank you very much for the help. at last I figure it in this way,
    the integration [itex]\int[/itex][itex]^{a}_{b}[/itex] f(x)dx can be viewed as summation of infinite zeros [itex]\underbrace{0+0+0+...}_{infinite}[/itex].
    But the form [itex]\underbrace{0+0+0+...}_{infinite}[/itex] is indetermined. the result of the sum depends on how to interpret it in the limits. If i define it in this way,
    [itex]\underbrace{0+0+0+...}_{infinite}[/itex]=lim[itex]_{N->\infty}[/itex][itex]\sum[/itex][itex]^{N}_{n=1}[/itex]0, then the sum is 0. just like other indetermined form [itex]\frac{\infty}{\infty}[/itex], 0/0.
     
  9. May 31, 2014 #8

    phinds

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    No, it is NOT a summation of zeros. As Halls said (more gently), you can think of it that way if you really feel the need to, but that won't make it right. You are still defining n/infinity as a number (zero in your particular arbitrary case) but your thinking of it that way does not MAKE it that way. Also, your statement that indeterminate forms are equivalent to zero is absolutely incorrect and in fact it is inherently a contradiction in terms. How can it be BOTH indeterminate AND determined as zero? Doesn't make sense.
     
  10. May 31, 2014 #9

    micromass

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    Infinity is a tricky concept. There are a lot of theorems that say that you can substitute ##\infty## in an equation and get the right answer ##0##. It is something that is done all the time.

    However, there are also many counterexamples to this procedure. Most of these involve exchanging limit and sum or infinite sums.

    The moral of the story is that a naive substituting of infinity is not always allowed and that a rigorous development of mathematics is necessary. So you need to have rigorous theorems that you have to cite in order to get the result that you want.
     
  11. Jun 1, 2014 #10
    yes, infinity is really a difficult concept. when I consider the definition of the integration,
    [itex]\int[/itex][itex]^{a}_{b}[/itex] f(x)dx=lim[itex]_{N->\infty}[/itex] [itex]\sum[/itex][itex]^{N}_{i=1}[/itex]f(x[itex]_{i}[/itex])dx[itex]_{i}[/itex] (1),
    I use the [itex]\infty[/itex] as the number and substitute it into the left side of the equ. (1).
    and leads to,
    lim[itex]_{N->\infty}[/itex] [itex]\sum[/itex][itex]^{N}_{i=1}[/itex]f(x[itex]_{i}[/itex])dx[itex]_{i}[/itex]=[itex]\underbrace{0+0+0+...}_{\infty}[/itex] (2)
    the expression [itex]\underbrace{0+0+0+...}_{\infty}[/itex] is meaningless and indetermined. To avoid such a trouble I should keep this expression out of the mind and stop using the ∞ as a number and remember that, as D H said, lim[itex]_{N->\infty}[/itex] [itex]\sum[/itex][itex]^{N}_{i=1}[/itex]f(x[itex]_{i}[/itex])dx[itex]_{i}[/itex] is the limit (if it exists) of the partial sums as the number of terms gets ever larger.
    Thank you very much for your discussions !
     
  12. Jun 1, 2014 #11

    phinds

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    NOW you've got it :smile:
     
  13. Jun 3, 2014 #12
    Thanks, friend. the infinity can be approached, but never be reached.
     
  14. Jun 3, 2014 #13

    phinds

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    Yep ... much like being pure in thought and deed :smile:
     
  15. Jun 3, 2014 #14

    micromass

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    Well, to be fair, infinity can be reached. That is, there are certain mathematical theories that says what happens if you actually have infinity, instead of simply approaching it. However, limits are not part of this theory. Limits only give information about what happens if you approach infinity.
     
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