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Infinite Sum Question

  1. Jul 16, 2012 #1
    I wasnt sure which math forum to ask this in so im putting it here, apologies if its supposed to be elsewhere.

    Anyway, my question has to do with an infinite sum.
    Suppose we have a series:
    each term in the series is double the previous term starting with 1

    if we multiply this by the number 1 it will remain the same
    1+2+4+8+16+32+... = 1(1+2+4+8+16+32+...)

    also (2-1) = 1
    so we can say
    1+2+4+8+16+32+... = (2-1)(1+2+4+8+16+32+...)
    if we expand the RHS we get
    1+2+4+8+16+32+... = (2+4+8+16+32+64+...) - (1+2+4+8+16+32+...)
    1+2+4+8+16+32+... = (2+4+8+16+32+64+...) + (-1-2-4-8-16-32-...)

    the first part of the RHS has a set of all positive even numbers and the 2nd part of the set has a set of all negative even numbers
    all the even numbers will subtract to 0 and i'm left with
    1+2+4+8+16+32+... = -1

    what have I done wrong? I can't figure what and where the error is.
  2. jcsd
  3. Jul 16, 2012 #2

    Quick question for you.

    Nevermind >_<
  4. Jul 16, 2012 #3


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    You've treated the sum of an infinite set as a finite numberyou're saying a = infinity, b= infinity + 1, so a-b = -1, infinity +1 = infinity.
  5. Jul 16, 2012 #4


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    Hi Routaran! :smile:
    You've subtracted two infinities from each other (because you chose a sum that doesn't converge). :wink:

    Your method works fine if you choose eg 1 + 1/2 + 1/4 + …

    try it! :smile:
  6. Jul 16, 2012 #5
    thats actually very neat. both sides approach the number 2.
    I guess I have a followup question then

    why did multiplying by (2-1) give me the correct answer for a convergent series but not for a divergent series?
    I mean i'm multiplying by 1, shouldn't the "value" remain the same?
  7. Jul 16, 2012 #6


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    because (as coolul007 :smile: also pointed out) you're writing ∞ - ∞ !! :rolleyes:
  8. Jul 16, 2012 #7
    I still don't understand why :confused:
    i mean i get what you and coolul007 are saying, ∞ - ∞ = ∞ but i dont understand how it applies to my first example.

    if i was to say for example
    set of all natural numbers - set of all positive even numbers
    (1+2+3+4+5+...) - (2+4+6+8+....)
    its obvious, even to me, that the result is still infinite because i have a set of all positive odd numbers left over.

    but in the other example, the only difference is the number 1, all the other terms are identical. I don't follow exactly how to end up with an infinity :frown:
  9. Jul 16, 2012 #8


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    Nooo that's not necessarily true. ∞ - ∞ could be anything! It could be ∞, 0, 1, -∞ or any other finite constant.
    Anyway, doing arithmetic with infinities is bound for failure. There's a reason we use limits instead :wink:

    Are you sure? Because with some similar manipulations as you did above, we can change the answer.

    A = (1+2+3+4+5+...) - (2+4+6+8+....)
    = (1+2+3+4+5+...) - 1/2(1+2+3+4+....)
    = (1-1/2)(1+2+3+...)
    = 1/2*(2-1)(1+2+3+...)
    = 1/2*( (2+4+6+8+...) - (1+2+3+4+5+...) )
    = 1/2*(-A)

    But you said earlier that A is obviously equal to [itex]\infty[/itex], so then now I've manipulated it to be equal to [itex]-\infty[/itex], but at the same time I've shown that A= -1/2*A so does that mean A=0? It seems to have multiple answers, all at the same time.

    Infinities, don't mess with them :tongue:
  10. Jul 16, 2012 #9
    I think I am starting to get a better idea. Are you saying that the order in which I perform arithmetic matters when I am dealing with infinities and that I cannot just multiply or add any which way I want?
    if so, is there a right/wrong way to deal with situations like this, something analogous to BODMAS for example, that guides on exactly how to proceed?
  11. Jul 16, 2012 #10


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    Also, you are claiming after multiplying by (2-1) you have 2 sets that have a different number of elements, one set has one more member, a -1, where did it come from???
  12. Jul 16, 2012 #11
    i see what i was doing incorrect now in that first example
    when i said
    (2+4+8+16+32+...) - (1+2+4+8+16+...)
    i was incorrect to subtract the 2nd number in the 2nd term from the 1st number in the first term. what i should have been doing was 1st number from 1st term - 1st number from 2nd term

    so i get
    (2+4+8+16+32+...) - (1+2+4+8+16+...)
    = (2-1) + (4-2) + (8-4) + (16-8) + ...
    = 1+2+4+8+...
    and the equality is preserved.

    I was very surprised to learn that the order, even in addition, matters.
    Thank you for the education.
  13. Jul 16, 2012 #12
    Yes, that is surprising. But one can say that the order matters here because [itex]1+2+4+8+...[/itex] sums to infinity. But even if the sum was not infinity, even then the order matters. For example, [itex]1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...[/itex] is an infinite sum which can be proven to equal a finite number, but still order matters.

    Series for which order does not matter are called commutatively convergent series.
  14. Jul 16, 2012 #13
    To be honest, the sum still equals -1 in a specific sense. In mathematics, this is often called analytic continuation.

    The Taylor series of the function [itex]\dfrac{1}{1-x}[/itex] is exactly the geometric series where the powers of x are taken, starting from 0 (when the series exist.) This means that the function we gave analytically continues the geometric series. Even where the series is divergent, a value can be assigned to the sum using the analytic continuation. Note that this DOES NOT equal the value of the actual sum, but in numerous physical and mathematical applications; the analytic continuation becomes important.

    The series you gave has a common ratio of 2, hence x=2. Plugging that into the analytic continuation function yields -1, which means -1 can be treated as a finite value that can be assigned to the sum (it does not equal the sum.) In fact, the geometric series which only makes sense when [itex]|x|<1[/itex] is mapped to a meromorphic function with a simple pole at x=1 with residue -1.
  15. Jul 16, 2012 #14
    This is why math people like to define what they are talking about :)

    What you've given is basically a special-case proof of the infinite geometric series formula: a_0/1-r. Here, a_0 = 1, and r = 2 so we get -1. Of course, as you noticed, it can't apply here.

    Consider the case A = 1+1+1+1...
    A = 0+A = 0+1+1+1...

    Subtracting "term by term", we will get 0 = 1! Complete nonsense...

    This leads us to define infinite series as "limits of partial sums"

    Now suppose we have a series a_0+a_1+...

    We let A_0 = a_0, A_1 = a_0+a_1 etc... and supposing that the sequences A_n "settles" to some number (converges is the technical word), we let A = lim A_n as n->infinity.

    Let us define B in the same way. Now it turns out we CAN do term by term subtraction and arrive at A-B.

    Our subtracted series will end up being: (a_0-b_0) + (a_1-b_0) ... which defined in the same partial sums way, is lim (A_n-B_n) as n-> infinity. Since {A_n} and {B_n} are convergent sequences, this evaluates to A-B.

    Anyways I am sure that this was hard to understand since I didn't give any details, but it's part of the motivation for studying analysis.
  16. Aug 9, 2012 #15
    I think your misconception is that you treated infinity like an everyday real (or more generally complex) number. I know this may not look intuitive, but ∞-∞ is not 0, but rather it is an indeterminate form! And ∞+1=∞. So what you did above is rather like taking ∞-∞, which I've just told you is not as simple as 0.

    Other indeterminate forms are like 0/0, 0 times ∞, and ∞/∞. About the infinity + 1 thing I mentioned above, there is something like: if you have infinite number of hotel rooms and infinite people such that all rooms are occupied, and one more person comes along and wants a room. Solution: Get each occupant to move one room down, and surprisingly enough, nobody would be left without a room. (This does not consider you need infinite amount of time to get your message across)
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