if I say that all the real numbers will have their negative pair ,example 5 and -5,so similarly for infinity there will be -infinity which will sum to 0

@Munem Sahariar
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It is very rare for an equation to have 2 answers.

At the very most, it would be "Under different interpretations". For example, \( 1 + 2 + 3 + \ldots \) would be infinity in the usual calculus treatment of infinite sums, but could be \( - \frac{1}{12} \) under the analytic continuation of \( \sum n^s \). As such, can you clarify under which interpretation we get the answer of "\(\infty\)", and under which interpretation we get the answer of "undefined"?

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## Comments

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TopNewestwhich wiki, which section, please explain

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The absolutely convergent wiki that I linked to. Read through the entire wiki, esp since it sounds like you are unfamiliar with this concept.

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why doesn't it work?

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Read the wiki, esp the last section.

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Mr.Calvin Lin, could you please explain, what actually you want to convey.

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The sum of all real numbers is undefined. This is because the summation is not absolutely convergent, and hence we cannot assign a value to it.

Why does the naive approach of "pairing up \(x\) with \(-x\) to make the sum of infinitely many \( x -x = 0\)" not work?

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if I say that all the real numbers will have their negative pair ,example 5 and -5,so similarly for infinity there will be -infinity which will sum to 0

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That's a good thought. Under what scenarios can we do? (Maybe the question to answer first is: what are we doing here?)

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is it 0?

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It's \(\infty\).

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Can you justify why?

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At the very most, it would be "Under different interpretations". For example, \( 1 + 2 + 3 + \ldots \) would be infinity in the usual calculus treatment of infinite sums, but could be \( - \frac{1}{12} \) under the analytic continuation of \( \sum n^s \). As such, can you clarify under which interpretation we get the answer of "\(\infty\)", and under which interpretation we get the answer of "undefined"?

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No, it is not.

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