Timeline for Why is $ \frac{\pi^2}{12}=\ln(2)$ not true?
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| S Aug 24, 2019 at 3:59 | history | edited | Ivan Izmestiev | CC BY-SA 4.0 |
improved formatting
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| S Aug 24, 2019 at 3:59 | history | suggested | Lincon Ribeiro | CC BY-SA 4.0 |
improved formatting
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| Aug 23, 2019 at 20:47 | review | Suggested edits | |||
| S Aug 24, 2019 at 3:59 | |||||
| Jun 9, 2010 at 23:54 | comment | added | Harald Hanche-Olsen | (@Qiaochu: My apologies for misspelling your name.) | |
| Jun 9, 2010 at 20:48 | comment | added | Harald Hanche-Olsen |
@Quiochu: One version goes like this: $\prod(1+a_n)$ converges unconditionally if and only if $\sum|a_n|<\infty$. To say the product converges unconditionally is to say that it converges to a non-zero, finite limit which is independent of reordering (after you throw away any terms with $a_n=-1$.
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| Jun 9, 2010 at 19:02 | comment | added | Qiaochu Yuan | @Helge: my recollection is that this is only true under some additional hypothesis, e.g. that the sum of a_n^2 converges (so that one can apply the obvious proof by taking the logarithm). | |
| Jun 9, 2010 at 17:46 | comment | added | Max Lonysa Muller | Ok.. thank you Helge! I guess you can imagine I'm a bit disappointed that my flow of (il)logical arguments was interupted that quickly... but you're right, of course. Suppose we can write sin(x)/x as the product of two infinite (divergent) products. Would the rest of the 'proof' be correct? | |
| Jun 9, 2010 at 17:40 | comment | added | Robin Chapman | Yes, you treated divergent infinite products as convergent. You get the same sort of problems as you do with treating divergent series as convergent, like $$0 = (1 - 1) + (1 - 1) +\cdots=1-(1-1)+(1-1)\cdots =1.$$ | |
| Jun 9, 2010 at 17:38 | vote | accept | Max Lonysa Muller | ||
| Jun 9, 2010 at 16:50 | comment | added | Helge | The statements: - The product of (1 + a_n) converges - The sum of a_n converges are equivalent. So we know that prod (1 + a_n), where a_2n = (1 + x/n) and a_2n = (1 - x/n) does not converge unconditionally. So by reordering it can achieve any positive value by the same argument as for sums. Just rewrite product (1 + a_n) = exp(sum log(1 + a_n)) to show this. | |
| Jun 9, 2010 at 16:31 | comment | added | M.G. | See en.wikipedia.org/wiki/Infinite_product and en.wikipedia.org/wiki/Conditional_convergence | |
| Jun 9, 2010 at 16:21 | comment | added | Max Lonysa Muller | Could you please elaborate on that? Why is it necessary for these prodcucts to converge? | |
| Jun 9, 2010 at 15:58 | history | answered | Helge | CC BY-SA 2.5 |