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Infinite sum

  1. Sep 5, 2005 #1
    If I added all the numbers between 0 and 1, what do I get?
     
  2. jcsd
  3. Sep 5, 2005 #2

    LeonhardEuler

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    Limit doesn't exist. There are infinitely many numbers between .5 and 1, so the sum is more than [itex]\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...=\frac{1}{2}\infty=\infty[/itex].
     
  4. Sep 5, 2005 #3

    TD

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    It would depend on what 'numbers' you're talking about.
    If you're talking about the reals, the answer is indeed [itex]\infty[/itex]
     
  5. Sep 5, 2005 #4
    Yeah weird to think about I guess, but

    [tex] \sum^{\infty}_{i =1} \frac{1}{i} [/tex] is a divergent series, so it has to be true. Seems that it is not something easy to think about intuitively
     
  6. Sep 5, 2005 #5
    Either adding up "all" the reals or rationals between 0 and 1 will produce infinity
     
  7. Sep 5, 2005 #6
    It seems strange that for p=1, the p-series is divergent. Why is this so? I mean, it seems like each term will be smaller than the one before it, and [itex]\lim_{n\to\infty}\frac{1}{n}=0[/itex] so it (at least intuitively) should be convergent.

    Any thoughts on this?
     
  8. Sep 5, 2005 #7
    Its just a case of adding up a lot of drops of water. The way that these drops of water are divided they happen to have an infinite volume. Obviously if there is some way to add them up in a rational way there will be a border between rational and irrational. Another way to look at it is that a sum is an integral where dx or dy or whatever is not infinitely small, however if the integral makes no sense then obviously the sum which is just an overestimate (for all but finitely many points depending on where you start the sum and your step size) won't either. 1/x integrates to lnx which as you are probably ware has an infinite amount of area under its curve as x approaches infinity.
     
  9. Sep 6, 2005 #8
    I don't know if this is really adding to the discussion or not, but if you perform the integral test on the (harmonic) series, (using a as any constant, doesn't matter), you end up with

    [tex] \int^{\infty}_{a} dx \frac{1}{x} [/tex]

    when integrated gives

    [tex] \ln{x}|^{\infty}_{a} = \ln{\infty} - \ln{a} = \ln{\frac{\infty}{a}} [/tex]

    But infinity divided by a constant is just infinity, and the natural log of infinity isn't bounded either. So, I guess through the use of the integral test we can indeed establish that the series diverges. To be honest, I don't remember if there was in fact a reason given in my calc class as to why the integral test works (it's been some time), so perhaps someone might have a better, or more in depth answer as to why the harmonic series diverges, and therefore, adding all the real numbers between, say, 0 and 1 is infinite.
     
  10. Sep 6, 2005 #9

    shmoe

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    One way to see the divergence is to group the terms like so (this is a little sloppy how I'm dealing with the limit, but you can make everything precise):

    1+1/2+1/3+1/4+...

    =1+1/2+(1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+...1/16)+...

    >=1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+(1/16+...+1/16)+...

    =1+1/2+2*(1/4)+4*(1/8)+8*(1/16)+...

    =1+1/2+1/2+1/2+...

    so by adding enoiugh terms you can make the original series as large as we like.


    The integral comparison test works fine as well and shows that the divergence is logarithmic (though you might have guessed that from the above). The integral test can be loosely justified by drawing the graph of 1/x and drawing a rectangle of hieght 1 from x=1 to 2, a rectangle of height 1/2 from 2 to 3, a rectangle of height 1/3 from 3 to 4, etc. The area of the first n rectangles is the sum of the first n terms of the harmonic series, and contains the graph of 1/x and hence the harmonic series diverges at least as fast as log. By shifting the rectangles one unit left you get the rectangles below the 1/x graph, so they will both have the same order of growth.


    Incidently, if you have any uncountable set of positive real numbers, then you can find a sequence of (distinct) numbers in this set whose series diverges.
     
  11. Sep 6, 2005 #10

    arildno

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    If you include 0 and 1, then your sum would be 1. (I assume that with "numbers", you meant the integers).
     
  12. Sep 6, 2005 #11

    Zurtex

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    I'm not quite sure that would make too much sense given the way the poster phrased the question.

    But yes, it is quite important to define what type of numbers you are talking about before you can prove anything.
     
  13. Sep 6, 2005 #12

    arildno

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    I don't think this makes a lot sense, either.
     
  14. Sep 6, 2005 #13

    matt grime

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    Adding up any uncountably indexed set of (positive) real numbers must necessarily yields infinity; partition the unit interval (in this case) up into the disjoint intervals

    [1/n,1/(n-1))

    each if the sum were in any sense to be finite then there can only be a finite number in each class, but then the set is a countable union of finite sets, hence countable, contradicting the assumption we were summing an uncountable set of numbers.
     
  15. Sep 6, 2005 #14
    That makes sense. It just didn't seem divergent intuitively.

    Thanks.
     
  16. Sep 6, 2005 #15

    matt grime

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    Oh, you're the poster who wanted sums to converge exactly when the terms in the sum went to zero. If so then you should do p-adic analysis since that is what happens there (but note there are some strange things, because in 3-adic analysis the sequence 2, 22, 222, 2222, 22222, .. converges (and to something quite strange).
     
  17. Sep 6, 2005 #16

    HallsofIvy

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    The requirement that sequence an go to 0 in order that the sum
    [tex]\Sigma_{n=0}^{\infty} a_n[/tex] converge is a necessary condition, not a sufficient condition.
     
    Last edited: Sep 6, 2005
  18. Sep 6, 2005 #17
    I think thats a weak assumption. On the real scale and numbers between 0 and 1 clearly means any real number between the two if you ask me.


    Thanks for everyones input.
     
  19. Sep 6, 2005 #18

    matt grime

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    *cough* and equally clearly asking what the sum of all numbers between 0 and 1 is makes so much sense..

    1. what order are you summing them in and

    2. obviously it diverges and if you'd bothered to check if this question had been asked before in this forum (which it has, in some sense; there was a thread about uncountable indices and summation, if not two)
     
  20. Sep 6, 2005 #19
    Sorry I dont have a PhD in math, i dont know what an uncountable index is. I don't see why it wouldnt make sense, and I dont see why the order would matter.

    My question was answered, I said thanks, what more do you want?
     
  21. Sep 7, 2005 #20

    matt grime

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    The following set of numbers, {1,-1/2,1/3,-1/4,1/5,-1/6.....} can be summed in (at least) two different ways to give two different answers. Thus the order of the summation matters (first year undergraduate analysis in the UK, not sure what year in the US, and quite possibly highschool in other countries).

    Irrespective of the order issue, the set you want to sum over contains only positive terms, and one ordering of them would be to sum 1/2, 1/3, 1/4 etc which is known to diverge.

    We know how to add up a finite number of terms:

    x+y+z,

    we know how to take the limit to a set indexed by the natural numbers

    x_1+x_2+x_3+.....

    but how were you going to add up all the numbers in [0,1]? What limit were you going to take? This is why, to me, it "clearly doesn't make sense" since there is no indication given of how to actually start summing this series. (the infinite set of numbers between 0 and 1 cannot be indexed by the natural numbers and is thus uncountable, that is the definition of uncountable).

    You said, rudely in my opinion, that arildno made a weak assumption, hence my (rude) reply; I'd say he made the most sensible assumption of them all.
     
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