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Infinite Summation Confusion

  1. Jan 13, 2016 #1

    rollingstein

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    I've been reading a bit about the very intriguing summation [itex] \displaystyle \sum_{n=0}^{\infty} {n}[/itex] and it seems [itex]\frac{-1}{12}[/itex] is the result but apparently with a lot of subtleties and caveats.

    It is those that I am trying to understand now.

    At first reading it appeared totally incongruous to me the sum of this infinite set of positive integers turning out to be a finite value and even that a negative value. I thought this was one more of those trick proofs that show 1 = 0 or some such thing by a carefully masqueraded disallowed manipulation (e.g. a division by zero).

    But on more reading it does not seem this is a trick. If so would it be mathematically rigorous to write [itex] \displaystyle \sum_{n=0}^{\infty} {n} = \frac{-1}{12} [/itex] or is there a better, more exact, more nuanced way to state this fact?

    Second, in some places I see it been written that [itex] \displaystyle \sum_{n=0}^{\infty} {n} - \int_{0}^{\infty} x dx = \frac{-1}{12} [/itex]

    To my naive observation this seems a different result: It tells us something about the difference between a summation and an integral. So is this the correct result or is the original sum the correct result or are both results correct? (Seems hard for both results to be simultaneously correct for then the second integral must go to zero, which would be mighty odd. )

    I did read the very interesting Wikipedia article (https://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯) about this and it seems to clearly say that the summation [itex] \displaystyle \sum_{n=0}^{\infty} {n}[/itex] is divergent.

    If so are these results mathematically not right but just that it makes sense to use them as [itex]\frac{-1}{12}[/itex] in some Physics-ey context? I did read about regularization & re-normalization etc. but I'm still getting a bit confused.

    A lot of that seems so handwavey and lacks the clarity typical of Math & Physics where a certain result just is true or is not true. And hence I felt compelled to ask: As stated are these summations rigorously correct or not?

    Alternatively, if one does encounter in a problem one is solving the term [itex] \displaystyle \sum_{n=0}^{\infty} {n}[/itex] can one always replace it by [itex]\frac{-1}{12}[/itex] to get a physically valid result? If not always, when can one do this?

    A second alternative: Is it correct to say that since "infinities" are not real in the first place, the result itself is just a curiosity that arises out of a flaw in our mathematical language and any result one might get by using the result [itex] \displaystyle \sum_{n=0}^{\infty} {n} = \frac{-1}{12} [/itex] will not be falsifiable by a real experiment anyways. i.e. A real experiment cannot probe a conjecture involving an infinity?
     
    Last edited: Jan 13, 2016
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  3. Jan 13, 2016 #2

    Samy_A

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    The short answer is simple, ##\displaystyle \sum_{n=0}^{\infty} {n}=+\infty##.

    As the Wikipedia article you linked explains, one can perform all kinds of informal manipulations on the series to get other results.
    The one I like most is the one with the Riemann zeta function. But it really stops there, you can't just replace ##\displaystyle \sum_{n=0}^{\infty} {n}## by ##\frac{-1}{12} ##, unless you can give a alternative definition of what ##\displaystyle \sum_{n=0}^{\infty}## means.

    Whether infinities are real or not, the usual mathematical definition of ##\displaystyle \sum_{n=0}^{\infty}a_n=B## (B a real number) doesn't need infinity to be "real".
    What it actually means is:
    ##\displaystyle \forall \epsilon>0 \ \exists k \in \mathbb N:\ \forall m>k:\ |\sum_{n=0}^{m}a_n-B|<\epsilon##. No infinity as you can see.

    And similarly, when we write ##\displaystyle \sum_{n=0}^{\infty}a_n=+\infty##, we mean something like:
    ##\displaystyle \forall K>0 \ \exists k \in \mathbb N:\ \forall m>k:\ \sum_{n=0}^{m}a_n>K##
     
  4. Jan 13, 2016 #3

    rollingstein

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    Thanks! I guess, informal manipulations seem very unlike the rigorous process I'm used to seeing in math & hence my confusion.

    Great that you clarified that: ##\displaystyle \sum_{n=0}^{\infty} {n}=+\infty##

    This I can grok. It is a rigorous, exact statement. Additionally, it agrees with my mathematical intuition too.

    So, to ask a follow-up question, is there a mathematically rigorous way (using the notion of limits etc.), if any, to express any equivalence between ##\displaystyle \sum_{n=0}^{\infty} {n}## and ##\frac{-1}{12}## whatever it may be. My point is can the informal equivalence be converted to a formal notation?

    Obviusly, ##\displaystyle \sum_{n=0}^{\infty} {n} \ne \frac{-1}{12} ## . So then what is the correct expression in this context that yields ##\frac{-1}{12}##
     
  5. Jan 13, 2016 #4

    Samy_A

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    Well, you could use the Ramanujan summation.
    It is perfectly rigorous, as long as one doesn't confuse it with the usual definition used for infinite series.
     
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