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?

  • 22
    $\begingroup$ $\frac{1}{n-1}-\frac{1}{n}=\frac{1}{n(n-1)}$ $\endgroup$ Apr 21, 2010 at 18:38
  • 2
    $\begingroup$ I cannot help wonder what causes someone to discover such things by accident. Especially if the two formulas are evaluated some considerable time apart. (For more transparent clarity, Gjergji's essential insight might have been written $\frac{1}{v-1}-\frac{1}{v}=\frac{1}{v(v-1)}$.) $\endgroup$ Apr 21, 2010 at 18:59
  • 1
    $\begingroup$ The above products come from two different streams of thinking. The first came from thinking about $\sum_{n=2}^{\infty} -\zeta'(n)$ -- taking its exponential leads to the first product. The second product came from convergence questions related to $\prod_{m=2}^{\infty} \prod_{n=1}^{\infty} \sqrt[m^{n}]{1+\frac{1}{m^{n}}}$. $\endgroup$ Apr 21, 2010 at 20:08

2 Answers 2


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?

  • 1
    $\begingroup$ The original product came up from investigating $\sum_{n=2}^{\infty} -\zeta'(n)$, that is if you exponentiate that you get the first product. The second product arose from trying to generalize $\prod_{n=1}^{\infty} \sqrt[2^{n}]{1+\frac{1}{2^{n}}}$. The number that arose from the computation looked fishy, and I checked my notes, and was stunned to see many digits agreeing. After some more numerical testing and a bout of frustrated algebra I decided to throw it here. $\endgroup$ Apr 21, 2010 at 19:38

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.