1
$\begingroup$

Euler obtained the famous result $1+2+3+...= -1/12$. Is this the only result for $\Sigma_{n=0}^{\infty}n$ or have other results been derived, for instance in set theory?

$\endgroup$
  • 2
    $\begingroup$ Please do not call this a "result". It is (by itself) simply false. $\endgroup$ – Gerald Edgar May 11 '17 at 18:49
  • 2
    $\begingroup$ en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF $\endgroup$ – user5699 May 11 '17 at 20:09
  • $\begingroup$ @Gerald Edgar: This IS a result. This equality cannot be "true" or "false" until LHS is DEFINED. With the LHS properly defined, one can compute it. The more precise version of the question would be "what are the reasonable DEFINITIONS of 1+2+3+... which have been proposed?" . $\endgroup$ – Alexandre Eremenko May 12 '17 at 6:05
  • 1
    $\begingroup$ $\sum n$ is perfectly defined, it is most certainly not the definition of analytic continuation of the Riemann zeta function. Saying that $\sum n = -1/12$ is a bit misleading. In particular it is not Euler's result. $\endgroup$ – Slereah May 12 '17 at 10:08
  • 1
    $\begingroup$ It is defined as a divergent series in $\Bbb N^\Bbb N$ $\endgroup$ – Slereah May 12 '17 at 15:04
2
$\begingroup$

First let me explain that Euler in fact obtained this result since some comments to the original question seem to doubt that.

Leonhard Euler applied the $\zeta$ series to obtain the result, $\zeta(-1) = -1/12$ but he knew that summation of divergent series is doubtful, namely "the cause of great dissent among mathematicians of whom some deny and others affirm that such a sum can be found." See Ed Sandifer: How Euler did it. http://eulerarchive.maa.org/hedi/HEDI-2006-06.pdf

In fact Euler applied the series where it does not converge at all and hence does not converge to $-1/12$ either. But at his times a lot of undigested math was en vogue. So he assumed with Leibniz and the Bernoullis that $1-1+1-1+1-+... = 1/2$ and with Wallis that $1/3 < 1/2 < 1/1 < 1/0 < 1/-1$. All that is wrong. Obviously $1+2+3+...> 1$. And if something else is meant by "$+$", then it should be clearly indicated. Unfortunately Ramanujan and many modern mathematicians accept this confusing notation. On the other hand, Euler has obtained correct results from wrong assumptions like $ log2\infty - log\infty= log2 $ or $(1 + x/\infty)^\infty = e^x$.

Bernard Bolzano in his Paradoxien des Unendlichen, p. 44f denotes the sum of all natural numbers by $S^0$. Further he introduces $N^0$ what we would call the cardinal number today, taking one unit for every natural number. He claims that some (without quoting names) have put $N^0(N^0+1)/2 = S^0$. But he does not endorse this result. He only claims that the sum $S^0$ must be by far larger than $N^0$. If I remember correctly, Euler somewhere wrote $i(i+1)/2$ with $i$ for numerus infinitus as he often did. But I am not sure.

Cantor summed all natural numbers with the result $ 1 + 2 + 3 + ... \nu... = \omega$ in a letter of 1883 to Mittag Leffler (i.e., before he had introduced cardinal numbers). He obtained the same sum for the harmonic series. He emphasized however the order of the terms: $2 + 3 + ... + \nu+ ... + 1 = \omega + 1$. Whether $2$ is necessary as first term of the sequence is left open but obviously it is not necessary. For another order he states $1 + 3 + 5 + ... + (2\nu + 1) + ... + 2 + 4 + 6 + ... + (2\nu) + ... = 2\omega$

$\endgroup$
  • $\begingroup$ The question was not how Euler obtained this but what are the alternative definitions which lead to a different result. $\endgroup$ – Alexandre Eremenko May 12 '17 at 13:37
  • $\begingroup$ I know, but there are comments doubting that Euler derived this result. If the first part collects some more downvotes then I will delete it. $\endgroup$ – user5737 May 12 '17 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy