0
$\begingroup$

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? $$\sum_{n=1}^k n = \frac{k\cdot(k+1)}{2}$$ $$\frac{\infty\cdot(\infty+1)}{2} = -\frac{1}{12}$$ $$\infty\cdot(\infty+1) = -\frac{1}{6}$$

$\endgroup$
  • $\begingroup$ While this question is partially a duplicate of that, I think there are some other issues not tackled in that question that should be addressed. Additionally, I don't think the downvotes are warranted - this is not a 'bad question'. The asker has come across something they don't understand and is asking for clarification - if we close the question and ignore it, how does that help them understand it any better? $\endgroup$ – preferred_anon Mar 12 '14 at 11:09
  • $\begingroup$ check this out numberphile video explaining the -1/12. Also look at other videos in particular the latest numberphile talking about the Riemann zeta hypothesis (clay prize). $\endgroup$ – Chinny84 Mar 12 '14 at 11:13
  • 1
    $\begingroup$ Oh, who's to say what's real and what's not anyway? :-) $\endgroup$ – Lucian Mar 12 '14 at 11:25
  • 2
    $\begingroup$ For someone who is learning mathematics, I won't recommend to watch this Numberphile's video, because these explanations are quite misleading. $\endgroup$ – user10676 Mar 12 '14 at 11:28
  • 3
    $\begingroup$ What I reproach to this video is that the man forgot that the equation $S=4S+1/4$ has solutions $-1/12$ and $\pm \infty$ (which is critical). $\endgroup$ – user10676 Mar 12 '14 at 12:48
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.