11

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?

Tobias Kienzler
  • 6,609

2 Answers2

8

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

Tobias Kienzler
  • 6,609
Angela Pretorius
  • 10,200
  • (+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) – Gottfried Helms Aug 16 '13 at 07:28
  • 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 07:42
  • "Any linear summation method should have"...so is the property not part of the definition of "linear" here? – Nikolaj-K Aug 16 '13 at 07:47
  • 1
    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 08:09
  • 4
    @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
  • There are arguments that 2+2+2+2+2... = 0, actually... – CommaToast Jul 08 '17 at 06:58
  • You wrote: $\sum_{n=0}^\infty\frac1{n^0}=\zeta(0)=-\frac12$. This is wrong. If you start from $n=0$, the sum is $\zeta(1)=1/2$. If you start from $n=1$, the sum is $-1/2$. If you do not specify from where you start and write $1+2+3+4+...$, the answer can be any. – Anixx May 01 '21 at 16:13
0

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,

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

Semiclassical
  • 15,842
Anixx
  • 9,119
  • I'm sorry, but "Mathematica says so" is not really convincing – Tobias Kienzler Jan 09 '21 at 10:09
  • Yet this downvoted answer was correct, and the given answer is incorrect. The statement in the question about zeta(0) is false, because as this answer pointed out he starts the sum at 0 and not 1. – Gerben Feb 27 '21 at 14:40
  • @Gerben Oh, my bad, I didn't read this answer as "you accidentally started the sum at $n=0$ instead of 1." Sorry! – Tobias Kienzler May 02 '21 at 17:54