Is $1 + 2 + 3 + \dots = -\frac{1}{12}$ really true?

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I've read this strange result of the sum of all positive integers being $-\frac{1}{12}$. Is it really true? Does this also mean this is true? $$\sum_{n=1}^k n = \frac{k\cdot(k+1)}{2}$$ $$\frac{\infty\cdot(\infty+1)}{2} = -\frac{1}{12}$$ $$\infty\cdot(\infty+1) = -\frac{1}{6}$$

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The statement is not true in the sense that it is provable.

The statement is true in the sense that as a definition, it is consistent (not contradictory, not paradoxical).

Or more exactly, it is a special case of an generalized definition of summations. You are probably familiar with the definition

$$\sum_{n=1}^\infty f(n) = \lim_{k\rightarrow \infty} \sum_{n=1}^k f(n)$$

Sometimes that limit does not exist. It is "undefined". So there are different ways of extending the definition of a summation in these cases, such as Abel Summation or Cesàro summation. An extended definition is preferred to maintain certain properties:

  • The new definition must agree with the old definition when the old definition isn't undefined
  • Linearity: $\sum_{n=1}^\infty M\,f(n) + g(n) = M\, \sum_{n=1}^\infty f(n) + \sum_{n=1}^\infty g(n)$
  • Some rule allowing for the index to be changed, like $m = c(n)$, and possibly a property like $\sum_{n=1}^\infty f(n) = f(1) + \sum_{n=2}^\infty f(n)$

Under a generalized definition of a summation, $1 + 2 + 3 + 4.... = -\frac 1 {12}$. Now hopefully people will quit asking about this.

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The question cited as 'possible duplicate' covers why $\sum_{n=1}^{k}n$ may or may not be considered to be $-1/12$. However, there are some other important points to make:

$$\sum_{n=1}^{k}n=\frac{k(k+1)}{2} \implies \frac{\infty(\infty+1)}{2}=\frac{-1}{12}$$

This would only be true if $k$ were a positive integer (otherwise, your sum, and more importantly the proof of this formula, does not make any sense).

$$\frac{\infty(\infty+1)}{2}=\frac{-1}{12} \implies \infty(\infty+1)=\frac{-1}{6}$$

This would only be true if we could multiply infinite quantites by finite quantites and maintain equality, but this is not the case. To see that, consider this: "Does $2\cdot \infty = \infty$?". You should realise that the question doesn't make any sense.