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Jul
6
answered Sum: $1-2+3-4+5-6+…$
Jul
6
revised Extending 2-adic valuation to real numbers
added 810 characters in body
Jul
6
revised Extending 2-adic valuation to real numbers
added 238 characters in body
Jul
6
revised Extending 2-adic valuation to real numbers
added 238 characters in body
Jul
6
revised Extending 2-adic valuation to real numbers
added 223 characters in body
Jul
6
answered Extending 2-adic valuation to real numbers
Jun
29
revised Is there something between summation and integration?
added 499 characters in body
Jun
29
comment Is there something between summation and integration?
this is answering a very different question actually, but it might be of interest
Jun
29
answered Is there something between summation and integration?
Jun
28
comment How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
@AsafKaragila I specify this was done with zeta function regularization. Would it be more precise to ask for the analytic continuation $\zeta(0)$ ?
Jun
28
comment How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
Could you prove Ramanujan summation as Euler-Maclaurin to 1st order?
Jun
28
comment How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
hmm? that seems plausible that $\sum (0+2) = \sum (1+1)$ and yet $\sum (0+2) = 2 \sum 1$.
Jun
28
comment How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
@AsafKaragila this false statement has its own wikipedia article
Jun
28
revised How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
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Jun
28
asked How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
Jun
28
comment What we get if we add 1/2 infinite times
@Ant There is nothing wrong with this kind of experimentation. Euler did it, Physicists do it routinely (via their lack of rigour). He just needs to understand these series come with at the expense of associativity. We can't invoke $(a+b)+c = a + (b+c)$ infinitely many times and always get away with it. Watch his parentheses carefully.
Jun
28
comment What we get if we add 1/2 infinite times
@Ant his question seems to be whether these divergent series identities can be found via elementary rearrangements.
Jun
28
revised Proof $\{(x,y,z)|4x^2+9y^2+16z^2<1\}$ is an open set
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Jun
28
answered What we get if we add 1/2 infinite times
Jun
28
answered Proof $\{(x,y,z)|4x^2+9y^2+16z^2<1\}$ is an open set