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Is $1+2+3+4+5.... = -\frac{1}{12}$ self-contradictory ? I've heared much that $1+2+3+.... = -\frac{1}{12}$, although the fact that this series is diverging. I saw a proof of it by a physicist. In fact, I thought about it a little bit and found that I can derive a contradiction from this equation.

Suppose $x=1+2+3+.... $

So, $x=1+2+3+4+.... = 1 + ( 2+ 3 + 4 + 5 + ... ) $

$= 1+ ( [1 + 1] + [1+2] + [1+3] + [ 1+4] +... ) = 1+ (1 + [1+1] + [2+1] + [3+1] + ...)$

(I have moved the square paraenthese only here one step to the right )

$=1+ (1+2+3+4+5+...)= 1+x$

So, $x=1+x$

So, $0=1$, a contradiction.

This is not the only way to derive the contradiction, I can get a contradiction using another mathod.

My question is, If this equality holds in fact, How can we deal with my contradiction?

I heared that the equality is proved using zeta function which I know nothing about. Is that proof valid? and if yes, How come? I mean, it seems to be self-contradictory then How can we prove something false? this means that mathematics is not sound!( usning mathematical logic terms)

For me, the fault is that, We pre-supposed that this series has a value. In fact, it doesn't and here is the gap. Is that true?

FNH
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    Try searching for duplicates first? – qwr Sep 07 '14 at 20:26
  • Your arithmetic is incorrect in the middle - there's no reason why the two parenthetical quantities should be equal; one of them is 'infinitely larger' than the other. But this subject has been tackled again and again and again here on this site. See, e.g., http://math.stackexchange.com/questions/39802/why-does-123-dots-frac112/ – Steven Stadnicki Sep 07 '14 at 20:27
  • I read it before posting my question. but It didn't talk about my contradiction so I posted it. @qwr – FNH Sep 07 '14 at 20:27
  • @StevenStadnicki, Can you point out exactly where my argument go wrong? – FNH Sep 07 '14 at 20:29
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    Your manipulation is legitimate if you regard it as an integral with respect to the counting measure. But in that view, the value is $+\infty$. And $1+(+\infty) = +\infty$ is not surprising. – Daniel Fischer Sep 07 '14 at 20:37

2 Answers2

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The series does not have that as its value, so any attempts to equate the two sides will not result in a contradiction for the reason that they do not have any actual value. Instead what you have done is show that it cannot converge.

What's going on here is a process called analytic continuation, meaning that a different function is being evaluated here, one that agrees with $\sum_{n=1}^\infty{\frac{1}{n^s}}$ for all the values of $s$ for which the series converges, but is continuous on all of $\mathbb{C}$ (and, in particular, exists for every $s\in \mathbb{C}\setminus\{1\}$). This function, when evaluated at $s=-1$ has the value $-1/12$. This analytic continuation is also unique, meaning that if the series did converge, it would need to converge to this value.

It's very unfortunate that this is always regarded as some kind of unintuitive equality, and some youtube channels like Numberphile make it seem like it is an equality. However, it is important to stress that it is not an equality, and instead just the only possible value that it could be if it did converge (which it doesn't).

Hayden
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  • I've added a one step more to justify it. – FNH Sep 07 '14 at 20:32
  • @MathsLover I've edited the answer to explain why it still doesn't matter. – Hayden Sep 07 '14 at 20:35
  • @Hayden I hate to be a pedant but I believe it exists for all of $\mathbb{C}$ except $s=1$. That being said, I'm very glad you've emphasized the point of it being an analytic continuation (& the numberphile video has endlessly frustrated me!) – Jam Sep 07 '14 at 20:39
  • @Eul Can That the analytic continuation exists for all of $\mathbb{C}$ except $s=1$, or that $\sum_{n=1}^\infty{1/n^s}$ exists for all of $\mathbb{C}$ except $s=1$? – Hayden Sep 07 '14 at 20:42
  • @Hayden , I got your point but I want to read more about analytic continuation to see how this works in action (even for simple examples not necessarly this particular one) I tried to read Wikipedia article but It doesn't seem as a good resource. Do you recommend anything? – FNH Sep 07 '14 at 20:47
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    @MathsLover The problem is that a bit of complex analysis is necessary for understanding it best. Many texts introduce analytic continuation after they've gone through the bulk of an intro course in complex analysis (covering, for example, analyticity of complex-valued functions of a complex variable, Cauchy's Theorem, Cauchy's Formula, Residue Theorem, etc) – Hayden Sep 07 '14 at 20:50
  • Ok, so Do you recommend that I wait till take a course on complex analysis? (not that far, I think!) – FNH Sep 07 '14 at 20:52
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    Probably, if you want to study it rigorously. The short and simple of it is that the analytic continuation of a function which possibly has singularities is a way to create an analytic function that agrees with the original function where it is defined (where analytic means, roughly, that it is equal to its Taylor series). – Hayden Sep 07 '14 at 20:53
  • @Hayden I meant the analytic continuation due to the pole at $\zeta(1):$ :) – Jam Sep 07 '14 at 21:17
  • @Eul Can Yes, it does have a simple pole at $s=1$, for some reason I kept reading what you wrote as $-1$. My bad! XD – Hayden Sep 07 '14 at 21:20
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Short answer: the 'series' on the left hand side is only a symbol, for which the ordinary rules of arithmetic do not necessarily hold.

Example: the series

$$ 1-\frac{1}{2}+\frac{1}{3}-\cdots $$

converges (in the ordinary sense) to $\log 2$. But according to a celebrated theorem by Riemann, by rearranging the terms (which you can always do with a finite sum without changing the value) we can get any real value we want (or even a divergent series)!

Similarly, divergent series such as yours (for which a number of sophisticated extended summation techniques are available) do not necessarily obey things like commutativity or associativity.

Kim Fierens
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  • I've read that mathematicians of 17th century believed that laws of arithemetic holds for infinite series and called this a princible which I don't remember anything about. Do you know anything about that? – FNH Sep 07 '14 at 20:48
  • @MathsLover Basically all the rules of arithmetic hold for absolutely convergent series. Commutativity and all its consequences fail with conditionally convergent series. Almost everything fails with analytically continued series. – Ian Sep 07 '14 at 20:56
  • Yes, Euler did some pretty weird stuff with conditionally convergent and divergent series, and still managed to get correct answers! But perhaps this reflects more on his superior intuition than on his actual understanding of the concept of convergence (which was tenuous anyway in the 17th and 18th centuries). – Kim Fierens Sep 07 '14 at 21:05