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
–
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"?

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestwhich wiki, which section, please explain

Log in to reply

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

Log in to reply

why doesn't it work?

Log in to reply

Read the wiki, esp the last section.

Log in to reply

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

Log in to reply

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?

Log in to reply

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

Log in to reply

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

Log in to reply

is it 0?

Log in to reply

It's \(\infty\).

Log in to reply

Can you justify why?

Log in to reply

Log in to reply

Log in to reply

Comment deleted 3 months ago

Log in to reply

Log in to reply

Comment deleted 3 months ago

Log in to reply

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"?

Log in to reply

No, it is not.

Log in to reply