# Is the sum of natural numbers equal to $-\frac{1}{8}$? [duplicate]

I came across the following video on YouTube: Sum of all natural numbers (- 1/8).

Basically what happens is: \begin{align*} 1+2+3+\dotsb &= N \\ 1+(2+3+4)+(5+6+7)+\dotsb &= N\\ 1+9+18+27+\dotsb &= N\\ 1+9(1+2+3+4+\dotsb)&= N\\ 1+9N &= N \end{align*} and therefore $N=-\frac{1}{8}$.

This is directly in contradiction with the well-known result of $-\frac{1}{12}$.

What is the problem with this reasoning? Was this result discovered earlier? Is this a consequence of Riemann's Rearrangement Theorem? Thanks in advance.

This was a repost of my previous post because some people said it was a duplicate to "Why is the sum of natural numbers $-1/12$?"

• There is one right answer to that question, and infinitely many wrong ones. -1/8 and -1/12 simply happen to be two different wrong ones. – dxiv Aug 21 '16 at 23:51
• I agree that your question is not a duplicate. I would be interested to see if there is a regularization method that yields $1+2+3+\ldots=-\frac{1}{8}$. The method I know e.g. Riemann zeta regularization yield $1+2+3+\ldots=-\frac{1}{12}$. – Batominovski Aug 21 '16 at 23:52
• @Batominovski Just compose the grouping operation OP used with zeta regularization. – arctic tern Aug 21 '16 at 23:53
• How exactly? Writing $\frac{1}{1^s}+\frac{2+3+4}{2^s}+\frac{5+6+7}{3^s}+\ldots=1+9\big(\zeta(s-1)-\zeta(s)\big)$? – Batominovski Aug 21 '16 at 23:54
• @Batominovski A more interesting question, to my mind, is whether different ("formally valid") regularisation techniques can yield different finite "sums"? We all know that "naive" algebraic manipulations of divergent series can give convergent "sums" but the process is invalid and the answers are "totally wrong". But do "valid" regularisation techniques all converge to the same solution for a given series? In other words, is there a single "correct finite sum" for a divergent series? (as an aside: that's the most number of quote marks ("...") I've ever used in a single comment or answer! :) – Deepak Aug 22 '16 at 0:36

What is the problem with this reasoning?

You grouped terms together, and indeed performed infinitely many groupings of terms. Generally speaking, this can change the regularized value of a divergent series with respect to a fixed choice of summation method.

No idea.

Is this a consequence of Riemann's Rearrangement Theorem?

No, that theorem only applies to (conditionally) convergent series, not to divergent series.

Let us try to work with that kind of logic. Let $$N=1+2+3+\cdots,\ \ \ \ \ \ M=1+1+1+1+\cdots$$ Then $$N-M=0+1+2+3+\cdots=N.$$ So $M=0$. Now $$-1=-1+0=-1+M=-1+1+1+\cdots=1+1+1+\cdots=0.$$ Not particularly sound: this proves that all numbers are zero.

The problem arises from treating $N$ and $M$ as numbers. We can assign them a number if we want, but as the above shows the assignment will not be coherent with the usual laws of arithmetic.

As mentioned in artic tern's answer, this has nothing to do with rearrangements of conditionally convergent series.

In your basic calculus course you define what it means for a series (an infinite sum) to be equal to a real number. One also defines exactly when one can say that a series is equal to a number.

Under that definition, we can say that the series/sum $$1 + 2 + 3 + 4 + \dots$$ is divergent. This means that it is not equal to $-1/8$ or $-1/12$. On might even say that it doesn't exist (as a limit of partial sums).

Now, when people then say stuff like $1 + 2 + 3 + \dots = -1/12$ they will have a definition that explains exactly what this means. And that definition will be different from the standard definition from your standard calculus class. To understand more about how one can give meaning to saying things like this, I suggest that you browse the following questions that have already touched on this:

There are other questions that a relevant (as you can see from the list on the side).

Infinite sums that do not converge can be assigned meaningful values in different ways.

$1+2+3+\dots = \infty$ and $1+2+3+\dots = -\frac 1{12}$ are both meaningful in the appropriate context.

But calculating with hyperreals has no bearing on this issue except that it makes the first choice reasonable in some contexts.