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Some of diverging series is Cezaro- or Abel-summable. There are some good properties of Cezaro and Abel summation: (1) inserting finitely many zeroes does not change the value, (2) interchanging finitely many terms does not change the value, (3) it is linear, (4) and the value of Cezaro and Abel summation of the converging series is its limit.

From the wikipedia, I found out that assigning a good value to $1+1+1+\ldots$ or $1+2+3+\ldots$ is impossible since $$ 1+1+1+\ldots = S\\ 0+1+1+\ldots = S $$ and subtracting both sides, we get contradiction $$ 1=0. $$ Hence we must discard one of those properties to assign the value to those series.

Is there also a contradiction when one assigns $1+2+4+\ldots$ to $-1$ with those properties assumed?

Edit: I think many of people are misunderstanding my question. Assigning a value to $1+2+4+...$ should have stability and linearlity like Abel summation. As I mentioned, zeta regulariztion, which assigns $-1/2$ and $-1/12$ to $1+1+1+...$ and $1+2+3+...$ would not satisfy stability or linearlity. I know that it is natural to assign $-1$ to $1+2+4+...$, since it is true in $2$-adic and analytic continuation of geometric series also tells us that, but I cannot find any reference that this assignment assures linearlity and stability. Also I am also of course not interested in the finite summation.

J1U
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    It's the $2$-adic sum. – Angina Seng Aug 11 '18 at 13:24
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    I know it is convergent in 2-adic but then $1+1/2+1/4+\ldots$ will not be convergent. – J1U Aug 11 '18 at 13:31
  • That's an ad-hoc requirement / argument. Nowhere in your question did you use the convergence of $1 + 1/2 + 1/4 + \cdots$. – Kenny Lau Aug 11 '18 at 13:40
  • I stated the fourth rule so that the assignment of the value should be compatible with other converging series. What I am not sure is if it is possible to use 2-adic topology when computing $1+2+4+\cdots$ and use the standard topology when computing $1+1/2+1/4+\cdots$. – J1U Aug 11 '18 at 14:13

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Your argument shows no stable linear method can assign a finite sum to your first two series. As explained here, the stable linear method Euler summation obtains $\sum_{k\ge 0}2^k=-1$. You can, however, prove no totally regular method would work, by considering partial sums.

J.G.
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  • @poetasis One of those series is not like the others; don't confuse $n^a$ with $a^n$. – J.G. Aug 11 '18 at 13:59
  • @poetasis They're similar enough they'll not become finite under a totally regular method. The dissimilarity is the polynomial series you've considered also can't even get away with a linear stable summation. – J.G. Aug 11 '18 at 14:08
  • Does Euler summation gives linearlity and stability? I couldn't find the reference about this. – J1U Aug 11 '18 at 14:17
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    @JTU Follow that link, then find the paragraph that starts "The fact that". – J.G. Aug 11 '18 at 14:27