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Using zeta function regularization, the divergent series $1+1+1+1+...$ can be evaluated to yield $$1+1+1+1+1+...=\sum_{n=0}^\infty\frac1{n^0}=\zeta(0)=-\frac12.$$

But what is $2+2+2+2+...$ then? On the one hand, it should be twice as much, but on the other hand each $2=1+1$ so it could also be $-\frac12$ again. My gut feeling is that factoring out the $2$ of this divergent series is formally "less" valid than expanding the twos into sums of ones, but is the a sensible non-ambiguous answer?

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ans: $-1$

Any linear summation method should have $\displaystyle \sum_{n=0}^{\infty}a_nk=k\sum_{n=0}^{\infty}a_n$.

A regular linear stable summation method need not give the same answer when $2$ is rewritten as $1+1$ for infinitely many of the $2$s, considering that infinitely permuting terms in diverging series can alter the sum, and that in $1+2+4+8+...=-1$, setting $2=1+1, 4=1+1+1+1$ etc. would make the sum $-\frac{1}{2}$.

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  • $\begingroup$ (+1). Plus: one should perhaps enhance the "infinitely" in the partial sentence "...for infinitely many of the 2s..." because this is exactly the point and one should explicitely focus on this. (Just an opinion) $\endgroup$ – Gottfried Helms Aug 16 '13 at 7:28
  • $\begingroup$ That makes perfectly sense, thanks for the more intuitive $1+2+4+8+... \neq 1 + (1+1) + (1+1+1+1) + ...$ example $\endgroup$ – Tobias Kienzler Aug 16 '13 at 7:42
  • $\begingroup$ "Any linear summation method should have"...so is the property not part of the definition of "linear" here? $\endgroup$ – Nikolaj-K Aug 16 '13 at 7:47
  • $\begingroup$ The last paragraph is a little bit misleading, because no stable method can sum the series $2+2+2+2+\cdots$ or $1+1+1+1+\cdots$ at all. A more direct example of the same argument would be the series $1-1+1-1+\cdots$, whose Cesaro sum is $1/2$, versus the "rewritten" version $1+0-1+1+0-1+\cdots$, whose Cesaro sum is $2/3$. $\endgroup$ – Chris Culter Aug 16 '13 at 8:09
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    $\begingroup$ @ChrisCulter: or grouped as $(1-1)+(1-1)+(1-1)+\dots=0$. Regularization is very sensitive to rearrangement. $\endgroup$ – robjohn Aug 16 '13 at 11:13

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