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$ \newcommand{ifelse}[3]{ \left( \begin{cases} #1\text{ if }#2 \\ #3\text{ otherwise} \end{cases} \right) } $ A recent Numberphile video on 1+2+3+... has made this question ("Why?") popular again, as can be seen in the many related and duplicate questions.

But after puzzling a bit with this, it seems that the rules that they use in that video are inconsistent...

As far as I can see, they assume the existence of a sequence summation operator $\sum_n a_n$ which is just standard summation for convergent series, and which additionally obeys the following rules: \begin{align} & \sum_n (a\text{ with all zeroes removed})_n = \sum_n a_n \tag{0} \\ & \sum_n (c \times a_n + b_n) = c \times \sum_n a_n + \sum_n b_n \tag{1} \\ \end{align}

However, if I am not mistaken, these rules quickly lead to an inconsistency: \begin{alignat}{2} 0 &= \sum_n 1 - \sum_n 1 &&\text{(arithmetic, assuming $\;\sum_n 1\;$ exists)} \\ &= \sum_n 1 - \sum_n \ifelse{0}{n=0}{1} \quad &&\text{(inserting a 0 does not change the sum)} \\ &= \sum_n \ifelse{1}{n=0}{0} &&\text{(from (1))} \\ &= 1 &&\text{(convergent series)} \\ \end{alignat} Therefore my question is: Is there a set of consistent rules for a summation operator that allows us to compute $\;\sum_n n\;$ ("$1 + 2 + 3 + \dots$")? Does that work on all sequences, or do we have to exclude some (like perhaps $\;\sum_n 1\;$), and if so, which exactly?

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MarnixKlooster ReinstateMonica
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    Terence Tao has an old blog post that you might find informative. – Harald Hanche-Olsen Jan 24 '14 at 06:51
  • @BalarkaSen Well, Grandi's series isn't convergent at all! You're right that assuming Cauchy's definitions, nothing can be done with divergent series. But $1+2+3+4+\cdots$ is obviously divergent, so if we want to say anything interesting about it, we must consider broader definitions. And the various summation methods for divergent series do have consistent rules. We need only keep in mind that the rules are narrower and trickier to apply than the rules for convergent series. – Chris Culter Jan 24 '14 at 07:53
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    http://en.wikipedia.org/wiki/Divergent_series#Properties_of_summation_methods is exactly what you are asking for. I have a related answer here http://math.stackexchange.com/questions/709342/is-1-2-3-dots-frac112-really-true/709393#709393 . To your question, the step you cannot assume is the first, I am not aware of any extended definition that defines the sum of ones, although the difference might follow from linearity and geometric progress of zero so I'm not completely sure. – DanielV Apr 01 '14 at 23:59

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Yes, that Numberphile video is a mess, and has been roundly criticized on other math sites. The video uses bad math.

What is really going on is like the following. Let $f(x)=1+x+x^2+\dots$. We know that, when $|x|<1$, that $f(x)=\frac{1}{1-x}$. We also know, from complex analysis, that $f(x)$ as only one "analytic continuation, and so, since $\frac{1}{1-x}$ is analytic, that must be the continuation.

Then saying that $1+2+3+\dots = -\frac1{12}$ is akin to saying that $1+2+2^2+\dots = -1$. It is not really true.

In the case of $1+2+3\dots$, we start with the so-called zeta function:

$$\zeta(s)=1^{-s}+2^{-s}+3^{-3}\dots$$

With this definition, $\zeta$ is only defined for $s>1$, but the function has an analytic continuation, and $\zeta(-1)=-\frac{1}{12}$. That doesn't really mean that $1+2+3+\dots=-\frac{1}{12}$.

See this Scientific American blog post for a roundup and some links.

Thomas Andrews
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If we interpret the entire enterprise as computing $\zeta(-1)$, then the first rule comes from manipulating Dirichlet series, but with $n^{-s}$ written as just "$n$". So you may object to writing $$-4\times(1+2+3+4+\cdots) = 0-2(2)+0-2(4)+\cdots,$$ but you probably agree that for appropriate values of $s$, we have $$-\frac{2}{2^s}\times\left(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\right) = \frac{0}{1^s}-\frac{2}{2^s}+\frac{0}{3^s}-\frac{2}{4^s}+\cdots$$ and so $$\left(1-\frac{2}{2^s}\right)\times\left(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\right)=\frac{1}{1^s}-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\cdots.$$ Since this identity of convergent series holds for $\Re(s)>1$, we may choose to interpret both series by analytic continuation for other values of $s$, and the identity will still hold, practically by fiat. Now, unlike the series on the left, the alternating series on the right is Abel summable everywhere, which justifies all the remaining manipulations as operating on a power series instead, including the insertion of a single $0$.

(By contrast, inserting a single $0$ doesn't have the expected effect on a Dirichlet series; it doesn't just "move every term to the right". That's why you were able to get nonsense from $\sum 1$.)

To summarize, every step of the video comes from a valid operation on Dirichlet series or power series, but with the variable suppressed to make it look like an ordinary numerical series. Whether a given series converges or is Abel summable for some given range of parameters tells you which rules you may consistently apply.

Chris Culter
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