# Is $2 + 2 + 2 + 2 + … = -\frac12$ or $-1$?

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?

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 $-\frac{1}{2}$.
• That makes perfectly sense, thanks for the more intuitive $1+2+4+8+... \neq 1 + (1+1) + (1+1+1+1) + ...$ example – Tobias Kienzler Aug 16 '13 at 7:42
• 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 $2/3$. – Chris Culter Aug 16 '13 at 8:09
• @ChrisCulter: or grouped as $(1-1)+(1-1)+(1-1)+\dots=0$. Regularization is very sensitive to rearrangement. – robjohn Aug 16 '13 at 11:13