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Ramanujan's Summation says that the sum of all integers is -1/12... 1 + 2 + 3...=-1/12.

If we define group G to be group of all positive integers, then the group contains all positive integers. Since -1/12 is negative, and the group only contains positive integers, -1/12 is not an element of the group.

Therefore, Ramanujan's Summation is wrong.

Prefix-1
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2 Answers2

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Generalizations of the notion of "infinite sum" necessarily sacrifice some of the desirable algebraic properties of ordinary sums in order to attain greater generality. There's no "free lunch," so to speak.

The definition of "sum" under which the series you describe "sums" to $-\frac{1}{12}$ (clearly) does not have the property that a "sum" of positive numbers must be positive.

It's important to note that there's no "one true definition" of "sum," and that we may pick and choose which definition we are using for the purpose we're currently interested in. In some situations, it is useful to define "sum" such that Ramanujan's identity holds, even if it renders our notion of "sum" less ideal in other ways.

user3716267
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  • This answer is inadequate. Let G be the group of all positive integers, we can conclude G also contains all sums of all integers. – Prefix-1 May 08 '20 at 01:45
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    Whence can we conclude that? If we interpret "sum" as the group operation then it follows from the group axioms, but clearly Ramanujan summation is not the group operator on the addition group of positive integers (for starters, it's not even a binary operator). – user3716267 May 08 '20 at 01:50
  • Group G obviously contains all sums of all integers (trivially), with all integers being present & the group operation (composition) being +.
    • is "addition". "Addition" gives "sum". Do I really need to prove that?
     – Prefix-1 May 08 '20 at 01:54
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    At this point, I suspect you're trolling. – user3716267 May 08 '20 at 01:55
  • Elaborate. This is trivial. @user3716267 – Prefix-1 May 08 '20 at 01:56
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    Let us define "florp" to mean the generalization of the infinite sum through zeta-function regularization as it is typically defined.

    I'll now claim that the "florp" of $1, 2, 3, 4...$ is equal to $-1/12$.

    Do you still object?

     – user3716267 May 08 '20 at 01:58
  • Florp = "generalization of the infinite sum through zeta-function regularization" Sum = "the result of addition"

    Florp ≠ Sum. You can't just change the definition of sum into florp.

     – Prefix-1 May 08 '20 at 02:02
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    Unfortunately, in mathematics, definitions don't work that way. A "sum" is whatever everyone agrees to call a "sum," and it means different things in different contexts.

    Mathematical names are (to a point) arbitrary, and what we name our mathematical machinery has no bearing on the truth of our mathematical results.

    The statement that "a florp is not a sum" is vacuous. It's a sum insofar as people often call it a sum, which is exactly the degree to which anything else is a sum.

     – user3716267 May 08 '20 at 02:05
  • Ok. I understand. Your explanation is adequate now. Yes, inconsistent mathematics.... This bothers me. Does the inconsistency not bother you? – Prefix-1 May 08 '20 at 02:07
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    @Prefix-1 There is no inconsistency here. – user76284 May 08 '20 at 02:11
  • The inconsistency is bothersome more at some times than others - occasionally, mathematical notation does become extremely confusing due to the reuse (and abuse) of terminology - but that doesn't invalidate any of the mathematics, even if it's sometimes troublesome in practice.

    Remember that there are only so many well-known words to use, and if we used a different term every time we generalized a definition we'd rapidly either run out of words or else end up using hard-to-remember nonsense-words (see: "florp") for everything. It's a "no-win" situation.

     – user3716267 May 08 '20 at 02:17
  • So my question is fair (and shouldn't be downvoted), in that, depending on how you define "sum" -- either as infinite sum or finite sum -- my argument will be valid or invalid according on how you define "sum". Thank you, @user3716267 and I hope you can undownvote my question lol because others may be confused over the same thing. – Prefix-1 May 08 '20 at 02:29
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Your argument is invalid. To see why, consider the following:

“Ordinary summation says $$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots = \frac{\pi}{4}$$ But the rationals are closed under addition. Therefore ordinary summation is wrong.”

That a set is closed under addition (i.e. finite sums) does not imply it’s closed under summation (i.e. infinite sums).

Induction lets you prove properties of finite sums only.

user76284
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  • Thank you. This makes more sense than the other answer. This can show the difference between finite sums and infinite sums, if the definition of "sum" was ambiguous between "finite sum" and "infinite sum". – Prefix-1 May 08 '20 at 02:21
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    @Prefix-1 That’s right. You can think of it this way: $+$ is a function that takes two numbers and returns a number. $\sum$ is a function that takes a (possibly infinite) sequence of numbers and returns a number. See here and here for more details. – user76284 May 08 '20 at 02:29