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I seem to remember a proof by Euler, involving infinite series, which was really complex (for a maths hobbyist). I believe it was sent in a letter to someone, and that it ended up with $\pi = 0$ or something. I was just wondering if anyone has any idea what I'm talking about, or if it was someone else.

Thanks.

Oscar S
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  • I think you are confused with sin(2pi) = sin(0) or such ? – mick Oct 22 '14 at 21:40
  • I've heard of false proofs that show $1+2+3+\cdots=-1/12$ or an incorrect limit showing $\pi=2$, but I have never heard of a false proof that $\pi=0$. – Clayton Oct 22 '14 at 21:40
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    Take a Cricle. The ratio of the diameter of the circle and the circumference is pi. The circle looks like 0 therefore $\pi=0\ \square$ – Ali Caglayan Oct 22 '14 at 21:41
  • It's not either of those (I do quite like the -1/12 one) although it might be the incorrect limit of \pi=2. This was a proof that Euler knew was wrong but I think he wanted to see if the other person could figure out where? I'm not sure, but a link to the \pi=2 would be nice. – Oscar S Oct 22 '14 at 21:43
  • @alizter That's amusing but no thats not it :) – Oscar S Oct 22 '14 at 21:45
  • http://math.stackexchange.com/questions/12906/is-value-of-pi-4 Is what I had in mind. – Clayton Oct 22 '14 at 22:08
  • Again I like that one but its still not it... I believe Euler substituted some obscure value into a general infinite series and rearranged it (in a way you can't with infinite series) and somehow got pi=0 or 2 or something like that. – Oscar S Oct 22 '14 at 22:17
  • "Assume $0=1$, therefore for every natural number $n$, $0=n$ and so for every rational number $\frac pq=\frac0q=0$, take any sequence of rational converging to $\pi$, then this sequence is constant $0$ therefore its limit is $0$ as well. Therefore $\pi=0$." – Asaf Karagila Oct 23 '14 at 22:36

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$$e^{2i\pi}=1=e^0\quad=>\quad2i\pi=0\quad=>\quad\pi=\dfrac0{2i}=0.$$

Lucian
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