Is $2 + 2 + 2 + 2 + ... = -\frac12$ or $-1$?

Question

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

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?

Answer 1

ans: $-1$

Any linear summation method should have $\displaystyle \sum_{n=1}^{\infty}a_nk=k\sum_{n=1}^{\infty}a_n$.

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 $-\frac{1}{2}$.

Answer 2

This claim is wrong:

$$\sum_{n=0}^\infty\frac1{n^0}=-\frac12$$

These equalities are correct:

$$\operatorname{reg}\sum_{n=0}^\infty1=\frac12$$

$$\operatorname{reg}\sum_{n=1}^\infty1=-\frac12$$

You can check them in Wolfram Mathematica.

Sum[1,{n,0,Infinity},Regularization->Dirichlet]
Sum[1,{n,1,Infinity},Regularization->Dirichlet]

So,

$$\operatorname{reg}\sum_{n=0}^\infty2=2\operatorname{reg}\sum_{n=0}^\infty1=1$$