Infinite product experimental mathematics question. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T21:11:54Z http://mathoverflow.net/feeds/question/22088 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22088/infinite-product-experimental-mathematics-question Infinite product experimental mathematics question. deoxygerbe 2010-04-21T18:19:39Z 2010-04-21T21:55:08Z <p>A while ago I threw the following at a numerical evaluator (in the present case I'm using wolfram alpha)</p> <p>$\prod_{v=2}^{\infty} \sqrt[v(v-1)]{v} \approx 3.5174872559023696493997936\ldots$</p> <p>Recently, for exploratory reasons only, I threw the following product at wolfram alpha</p> <p>$\prod_{n=1}^{\infty} \sqrt[n]{1+\frac{1}{n}} \approx 3.5174872559023696493997936\ldots$</p> <p>(I have cut the numbers listed above off where the value calculated by wolfram alpha begins to differ)</p> <p>Are these products identical or is there some high precision fraud going on here?</p> http://mathoverflow.net/questions/22088/infinite-product-experimental-mathematics-question/22093#22093 Answer by Mariano Suárez-Alvarez for Infinite product experimental mathematics question. Mariano Suárez-Alvarez 2010-04-21T18:31:07Z 2010-04-21T18:31:07Z <p>An experimental observation: if $a_r=\prod_{v=2}^{r} \sqrt[v(v-1)]{v}$ and $b_r=\prod_{n=1}^{r} \sqrt[n]{1+\frac{1}{n}}$, then the numbers $a_{2r}/b_{2r}$ are, according to Mathematica, $$\frac{1}{\sqrt{3}},\frac{1}{\sqrt[4]{5}},\frac{1}{\sqrt[6]{7}},\frac{1}{\sqrt[4]{3}},\frac{1}{\sqrt[10]{11}},\frac{1}{\sqrt[12]{13}},\frac{1}{\sqrt[14]{15}},\frac{1}{\sqrt[16]{17}},\frac{1}{\sqrt[18]{19}},\frac{1}{\sqrt[20]{21}},$$ $$\frac{1}{\sqrt[22]{23}},\frac{1}{\sqrt[12]{5}},\frac{1}{3^{3/26}},\frac{1}{\sqrt[28]{29}},\frac{1}{\sqrt[30]{31}},\frac{1}{\sqrt[32]{33}},\frac{1}{\sqrt[34]{35}},\frac{1}{\sqrt[36]{37}},\frac{1}{\sqrt[38]{39}},\frac{1}{\sqrt[40]{41}},\dots$$ I would imagine the products are the same, then. I don't have time but using this as a hint one should be able to give an actual proof.</p> <p>May you tell us how you ended up with such an identity?</p> http://mathoverflow.net/questions/22088/infinite-product-experimental-mathematics-question/22107#22107 Answer by deoxygerbe for Infinite product experimental mathematics question. deoxygerbe 2010-04-21T19:57:20Z 2010-04-21T19:57:20Z <p>Aha, I get Gjerji's insight, and I should have seen it sooner, but I was stuck on dealing with series representations by logarithms. </p> <p>The second product looks like this:</p> <p>$\sqrt[1]{\frac{2}{1}}\sqrt[2]{\frac{3}{2}}\sqrt[3]{\frac{4}{3}}\sqrt[4]{\frac{5}{4}}\ldots$, and it can be rewritten like this: $2^{1-\frac{1}{2}}3^{\frac{1}{3}-\frac{1}{4}}4^{\frac{1}{4}-\frac{1}{5}}5^{\frac{1}{5}-\frac{1}{6}}\ldots$, and then one can employ Gjerji's observation to convert the second into the first.</p>