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# Sum

What is the sum of all real numbers?

Note by Adharsh M
7 months ago

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which wiki, which section, please explain

- 7 months ago

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

Staff - 7 months ago

why doesn't it work?

- 7 months ago

Read the wiki, esp the last section.

Staff - 7 months ago

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

- 7 months ago

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?

Staff - 7 months ago

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

- 7 months ago

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

Staff - 7 months ago

is it 0?

- 7 months ago

It's $$\infty$$.

- 7 months ago

Can you justify why?

Staff - 7 months ago

Because there are infinitely many real numbers.

- 7 months ago

And so?

Staff - 7 months ago

Comment deleted 3 weeks ago

So, you are changing your answer from "$$\infty$$" to "undefined"?

Staff - 7 months ago

Comment deleted 3 weeks ago

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"?

Staff - 7 months ago

No, it is not.

Staff - 7 months ago