Infinite product experimental mathematics question. - MathOverflow

Infinite product experimental mathematics question.

2010-04-21T18:19:39Z

A while ago I threw the following at a numerical evaluator (in the present case I'm using wolfram alpha)

$\prod_{v=2}^{\infty} \sqrt[v(v-1)]{v} \approx 3.5174872559023696493997936\ldots$

Recently, for exploratory reasons only, I threw the following product at wolfram alpha

$\prod_{n=1}^{\infty} \sqrt[n]{1+\frac{1}{n}} \approx 3.5174872559023696493997936\ldots$

(I have cut the numbers listed above off where the value calculated by wolfram alpha begins to differ)

Are these products identical or is there some high precision fraud going on here?

Answer by Mariano Suárez-Alvarez for Infinite product experimental mathematics question.

2010-04-21T18:31:07Z

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.

May you tell us how you ended up with such an identity?

Answer by deoxygerbe for Infinite product experimental mathematics question.

2010-04-21T19:57:20Z

Aha, I get Gjerji's insight, and I should have seen it sooner, but I was stuck on dealing with series representations by logarithms.

The second product looks like this:

$\sqrt[1]{\frac{2}{1}}\sqrt[2]{\frac{3}{2}}\sqrt[3]{\frac{4}{3}}\sqrt[4]{\frac{5}{4}}\ldots$, 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.