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If this is correct, then the demonstration below must have a fault, but I can't find it.

Assuming $1+2+3+\cdots = -\frac{1}{12}$ (1) is true.

Adding $0$ on both sides: $0 + 1+2+3+\cdots = 0 -\frac{1}{12}$ (2)

Subtracting (2) from (1),

we get, $1+1+1+1+1+1+\cdots=0$ (3)

Adding $0$ on both sides again, $0+1+1+1+1+1+\cdots = 0 + 0$ (4)

Now subtracting (4) from (3)

we get, $1=0$, which is a contradiction.

Obviously our assumption that $1+2+3+4+\cdots = -\frac{1}{12}$ is wrong.

There is some error in the steps above, but I cannot find it. What is it?

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Mindwin Remember Monica
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    $1+2+\cdots$ and $0+1+2+\cdots$ do not converge, so they are not real numbers, so you cannot use them in basic operations such as subtractions. Subtracting (2) from (1) you obtain $\infty-\infty$ on the left hand side, which is undefined (and definitely not defined as $1+1+\ldots$). – LinAlg Oct 25 '16 at 20:41
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    It is not actually true that $1+2+3+\cdots = -\frac{1}{12}$ for the standard definition of infinite sums. There are just some contexts in which you get useful results if you treat it as being true. – Eric Wofsey Oct 25 '16 at 20:45
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    Did you read all of the original question, its context, the tags for the question, all the comments, the comments challenging the answer you link to, and all the other answers offered? – amWhy Oct 25 '16 at 21:09
  • Essentially, https://twitter.com/ZachWeiner/status/625711339520954368 – Asaf Karagila Oct 25 '16 at 22:28

1 Answers1

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The error is that you assumed a result that is false when using standard definitions of summation and convergence.

The result $1+2+3+\cdots = -\frac{1}{12} $ is only true when nonstandard methods of summation are allowed.

This is like claiming that $1+2+4+8+... =-1 $ since $\dfrac1{1-x} =1+x+x^2+... $ and substituting $x=2$ to gives $1+2+4+... =\dfrac1{1-2} =-1 $.

marty cohen
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