1

In Prof Norman Wildberger's A Socratic look at the logical weaknesses of modern pure mathematics (which just made available on youtube), he mentioned a discovery by Euler (30:55) that:

$$...x^{-3}+x^{-2}+x^{-1}+1+x^1+x^2+x^3.. = 0$$

I'd like to learn more about it (I'm especially interested in Euler's approach to this) so I try googling but without the right keywords I can't find anything.

Here is my take in this:

$$S = ... x^{-3}+x^{-2}+x^{-1}+1+x^1+x^2+x^3 ... \\ x \cdot S = ... x^{-2}+x^{-1}+1+x^1+x^2+x^3+x^4.. = S $$

$$\forall x \in \mathbb{R} (x \cdot S = S \Leftrightarrow S \in \mathbb{R}) \Leftrightarrow (S =0) $$

with this I conclude that $S=0$ iff $S$ is a real number.

But what are some other more interesting ways of giving a sum to this divergent series? How did Euler do it? Is there a name for this special power series? Where can I learn more about it?

Update: perviously I have used the word "prove" but it appears that this is not exactly the "right" word to be used here so I have changed it to "giving a sum to a divergent series" ( e.g. in the case where $1+2+3+4... = -\frac{1}{12}$ ).

Archy Will He 何魏奇
  • 883
  • How is this supposed to hold for any $x > 0$? All summands are strictly positive, so their sum surely can't be $0$? – Huy Aug 07 '15 at 06:38
  • 1
    Note that $S=\infty$ also "satisfies" $x \cdot S = S$... – angryavian Aug 07 '15 at 06:39
  • @angryavian Yeah. I should have restricted $S \in \mathbb{R}$ – Archy Will He 何魏奇 Aug 07 '15 at 06:42
  • It's one of "those" series that you have look at in a regularized way. Anyway, you can take it as two geometric series: $1/(1-x)+1/(x(1-1/x))=0$. But of course, the first series converges for $|x|<1$ and the second for $|x|>1$ so you need to be careful how you define the sum. – orion Aug 07 '15 at 06:44
  • 'Proved' is a bit of an overstatement. The series $\sum_{n=-\infty}^\infty x^n$ does not converge for any $x$ so the result $S=0$ does not hold in the normal sense. – Winther Aug 07 '15 at 06:44
  • 1
    Euler discovered this? Baloney.It's a non-existent sum. You may be able to assign it a value,but not with the usual definitions for series and limits. If Mr.N.W. thinks otherwise then he is full of it. – DanielWainfleet Aug 07 '15 at 06:47
  • 1
    Anecdotally (disclaimer: I am not a mathematical historian), Euler did lots of things with infinite series that do not stand up to modern ideas of rigour. (At least) some of his results are in fact true. If Euler "proved" the above, I am sure he realised that the answer is not in fact correct. I am also sure that Norman Wildberger is not asserting that the sum really is zero, but is making a point about standards of proof in mathematics. – David Aug 07 '15 at 06:51
  • @Winther True, I should have used the phrase "giving a sum to a divergent series". Regarding that I have updated my question. – Archy Will He 何魏奇 Aug 07 '15 at 06:53
  • @David Sorry if I have used the wrong word. When it comes to proof I believe it all depends on the axioms the system is based on. I am just interested in his approach in giving a sum to this divergent series. – Archy Will He 何魏奇 Aug 07 '15 at 06:57
  • I asked a question about that subject..: . I got also the same result as euler gotvia using series solution in first method. Please see my question . http://math.stackexchange.com/questions/1359401/is-sum-n-infty-infty-xn-0 – Mathlover Aug 07 '15 at 07:07
  • @Mathlover thanks for the link. It's useful :) – Archy Will He 何魏奇 Aug 07 '15 at 07:24

0 Answers0