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I've seen methods to calculate this sum - also in questions on this site. But it seems it is a matter of how you want to regularize the problem. Are there summation methods which could give a different, finite result for this sum?

EDIT: One answer points out a simple solution. Apparently Wikipedia already mentions that transformations assuming linearity and stability will lead to inconsistencies (http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#Failure_of_stable_linear_summation_methods) Is there a derivation with less assumptions about the divergent series still giving a different result?

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    $\begingroup$ They would all be based on some abuse of mathematics since $1+2+3+\cdots$ apparently cannot be finite? $\endgroup$ – mathse May 3 '14 at 10:15
  • $\begingroup$ Sure. But with additional assumptions, apparently you can get an answer that might be even useful. So the question is whether there are two contradictory, but sensible assumptions possible. $\endgroup$ – Gerenuk May 3 '14 at 10:25
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Using the same zeta regularization that is used to get $$ 1+2+3+\dots=-\frac1{12}\tag{1} $$ we get $$ 1+1+1+\dots=-\frac12\tag{2} $$ Subtract $1$ from $(1)$ to get $$ 2+3+4+\dots=-\frac{13}{12}\tag{3} $$ Subtract $(2)$ from $(3)$ to get $$ 1+2+3+\dots=-\frac7{12}\tag{4} $$ Regularization of divergent series can lead to contradictions.

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  • $\begingroup$ That is fantastic! Now you have $1+2+3+\cdots = -\frac{7}{12}$ but also $1+2+3+\cdots=-\frac{1}{12}$. So adding $1+2+3+\cdots$ to itself and factoring out $2$ yields $1+2+3+\cdots = -\frac{1}{3}$ - and this is not yet the end of the story. $\endgroup$ – mathse May 3 '14 at 10:35
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    $\begingroup$ @mathse: yes, by adding $(1-x)\left(-\frac1{12}\right)+x\left(-\frac7{12}\right)=-\frac1{12}-\frac x2$ one can get the sum to be anything one wants. $\endgroup$ – robjohn May 3 '14 at 10:37
  • $\begingroup$ The wikipedia page says, that of course this series cannot be treated as linear and stable. Are there any derivations which don't use these two properties? $\endgroup$ – Gerenuk May 3 '14 at 10:45
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    $\begingroup$ The result/derivations depend largely on the assumptions you make. And from Wikipedia it seems if you use both linearity and stability (as in your case), you easily get contradictions. However, they mention two methods which use only one of the properties and yet both get $-1/12$. So, will all methods using linearity XOR stability get the same result? This seems more interesting. $\endgroup$ – Gerenuk May 3 '14 at 11:59
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    $\begingroup$ @Gerenuk: zeta regularization is stable but not linear. So actually the only property used to get a contradiction is linearity. $\endgroup$ – robjohn May 3 '14 at 13:21
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The answer depends on what you consider a method of summation. There are many, with the "classic" one given by convergence of partial sums being only a "foundation stone" - and of course this one doesn't assign a (finite real) value to $\sum n$. The various methods may not only assign different values to certain series, they may have different opinions whether or not a certain sum is defined in the first place. There are certain properties a summation method may or may not have, which allow us to speak very generally about them:

  • Regularity: A summation method is called regular if it coincides with the series whenever the latter converges. That is, if $\{a_n\}_{n=1}^\infty$ is a sequence such that $\sum_{n=1}^N a_n$ converges to some number $A\in\mathbb R$ as $N\to \infty$, then $\sum a_n=A$.
  • Linearity: If $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ are sequences such that $A:=\sum a_n$ and $B:=\sum b_n$ are defined and $u,v$ are real numbers, then $\sum (ua_n+vb_n)$ is defined and its value is $uA+vB$.

  • Stability: If $\{a_n\}_{n=1}^\infty$ is a sequence and we define $b_n=a_{n+1}$ then if one of $A:=\sum a_n$ and $B:=\sum b_n$ is defined, then so is the other and the equation $A=a_1+B$ holds.

A well-know example of a summation method is Cesàro summation: Let $s_N=\sum_{n=1}^Na_n$ and define $\sum a_n = \lim_{N\to\infty}\frac{s_1+s_2+\ldots +s_N}{N}$ whenever the limit exists. This summation method is regular, linear, and stable. And it is able to handle the case $a_n=(-1)^n$, so it is a proper extension of classic series summation!

Apart from Cesàro summation there are many othes, but interestingly they often agree on many results. In fact, certain results can be derived merely by the properties listed above. For example, if a summation is linear and stable (but not necessarily regular) and assigns a value to the geometric series $G(x)=\sum_{n=1}^\infty x^n$ for certain $x$, then it must assign a very specific value. Indeed we find by using statbility and linearity that $$G(x)=\sum_{n=1}^\infty x^n = x+\sum_{n=1}^\infty x^{n+1}=x+x\cdot \sum_{n=1}^\infty x^{n}=x(1+G(x))$$ thus necessarily we have $G(x)=\frac x{1-x}$ if $x\ne 1$ and $G(x)$ is defined. Specifically, $G(-1)$ is either $-\frac12$ or undefined for any stable linear summation (and indeed this is also the value Cesàro computes). We also see that if $G(1)$ is defined for a stable linear summation then $G(1)=1+G(1)$; we conclude that $G(1)$ is not defined for any stable linear summation.

As a corollary, $S:=\sum n$ is not defined for any stable linear summation. Indeed, if it were, we'd find by regularity that $\sum (n+1)$ is also defined and equals $-1+S$. Then by liniearity $$-1=(-1+S)-S=\sum(n+1)-\sum n=\sum 1 $$ would be defined as well.

"Unfortunately", stabilty and linearity feel like vary desireable properties of a summation if it is to somehow comprise the notion of sum of the terms at all. So when looking for summations that assign a value to $\sum n$ we have to throw overboard at least one of these conditions. But as long as we do not somehow restrict what we consider a summation method and what not, there is no obstacle in assigning an arbitrary value to $\sum n$. For example, we could declare $\sum a_n=P(42)$ whenever there exists a polynomial $P(X)\in\mathbb R[X]$ such that $\sum_{n=1}^N a_n-P(N)$ converges to $0$ as $N\to\infty$. This summation would even be regular and linear (but not stable). Since $\sum_{n=1}^Nn=\frac{N(N+1)}{2}$, it would assign $\sum n= 903\ne-\frac1{12}$.

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An interesting analogue is that it makes perfectly good sense in the $2$-adic integers $\mathbb{Z}_{2}$ that the series $1 + 2 + 4 + \ldots = \sum_{n=0}^{\infty} 2^{n} = -1.$ We can see that in any integral domain $R$ in which the series $\sum_{n=0}^{\infty} 2^{n}$ exists, and $ = \ell,$ say, we must have $\ell = -1.$ For $\ell - 2 \ell = 1$ in $R.$

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