I feel like this question has been asked before, but I couldn't find any question about this, so therefore if you find the same question somewhere(or even similar) please close this one.

My question is whether there exists some formula for calculating:

$$\prod_{k=2}^n \ln(k) = \ln(2)\ln(3)\ln(4)\cdots\ln(n)$$

My guess is that there probably isn't a formula (I tried to "doodle" around with it for a while and couldn't find anything), but if there is, please tell me :)

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    $\begingroup$ There definitely is nothing more simple and exact, but you could probably find asymptotics. $\endgroup$ – Raoul Apr 11 '20 at 18:21
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    $\begingroup$ It will be interesting to see what you get if you use complex numbers representation. $\endgroup$ – Moti Apr 11 '20 at 19:12
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    $\begingroup$ Since $\ln(a^b)=b\ln(a)$, we have that $\ln(a)\ln(b)=\ln(a^{\ln(b)})$, and since $\ln(b)\ln(c)=\ln(b^{\ln(c)})$, we have that $\ln(a)\ln(b)\ln(c)=\ln(a^{\ln(b^{\ln(c)})})$ and so on. Thus we have $$\ln(2^{\ln(3^{\ln(4)...^{\ln(n)}})})$$ I'm not sure where to take it from here. $\endgroup$ – 4yl1n Apr 14 '20 at 15:44
  • $\begingroup$ @4yl1n Yes, I'm aware of this fact, but I also don't know where to take it from there :(. Another try is to somehow group the terms, e.g. so that you group the powers of numbers that $$(\ln(2)\ln(4)\cdots\ln(2^a))(\ln(3)\ln(9)\cdots\ln(3^b))\cdots$$ This could be simplified further by bringing down the exponents so that you you would get $$a!\ln^a(2)b!\ln^b(3)\cdots$$ but this gets really complex so I don't know. Anyways, thanks for your help! $\endgroup$ – Casimir Rönnlöf Apr 14 '20 at 16:04

$$\ln(a)\ln(b)\ln(c)...=\exp(\ln(\ln(a)))\exp(\ln(\ln(b)))\exp(\ln(\ln(c)))...$$ (This is because $x=\exp(\ln(x))$, so applying this for every $\ln(a), \ln(b), \ln(c), ...$ yields the above formula).

Since $\exp(a)\exp(b)\exp(c)...=e^ae^be^c...=e^{a+b+c+...}=\exp(a+b+c+...)$, we can rewrite this as $$\exp(\ln(\ln(a))+\ln(\ln(b))+\ln(\ln(c))+...)$$

If we apply this to your question and use sigma notation, we get

$$\exp\sum_{k=2}^{n}{\ln(\ln(k))}$$ which cannot be further simplified (as far as I know) without going back to the original expression. So the answer to your question is no, unless you like $\exp\sum_{k=2}^{n}{\ln(\ln(k))}$.


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