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? $$\sum_{n=1}^k n = \frac{k\cdot(k+1)}{2}$$ $$\frac{\infty\cdot(\infty+1)}{2} = -\frac{1}{12}$$ $$\infty\cdot(\infty+1) = -\frac{1}{6}$$
Is $1 + 2 + 3 + \dots = -\frac{1}{12}$ really true?
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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.
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:
Under a generalized definition of a summation, $1 + 2 + 3 + 4.... = -\frac 1 {12}$. Now hopefully people will quit asking about this.