Asymptotics for $ f(n) = \prod_{k=2}^{n} \sqrt[k-1]{k}$?

Question

I was thinking about this function

$$f(n) = \prod_{k=2}^{n} \sqrt[k-1]{k}$$

Maybe use it to do number theory or so.

But then I started to wonder about the asymptotics of $f(n)$.

I first assumed $g(n) = n \ln(n)^{\ln(n)^{\frac{10}{9}}}(n) $ to be a good asymptotic.

But

$$\lim \frac{f(n)}{g(n)} = \infty$$

Then I considered $h(n) = \exp(\ln(n)^{\ln(\ln(n)/\sqrt \ln(n))})$

But again

$$\lim \frac{f(n)}{h(n)} = \infty$$

Maybe I should use different type of functions to make asymptotics? Gamma?

Or do I need more complicated special functions to get a good asymptotic like tetration or Ackermann function?

I know $\ln(n) = \lim_m m(n^{1/m}-1)$ so maybe I should use $\ln(n!)$ as part of the asymptotic?

This also implies looking at

$$t(n) = \prod_{k=2}^{n} \frac{\sqrt[k-1]{k}}{\ln(k+1)}$$

or

$$v(n) = \prod_{k=2}^{n} \frac{\sqrt[k-1]{k}}{\ln(k+1)^\dfrac{1}{\ln(k+1)^2}}$$

But those 2 do not seem to converge.

Should I try Taylor series?

How to get a good asymptotic? Does it even exist?

Answer 1

By the Abel–Plana formula \begin{align*} \sum\limits_{k = 1}^{n - 1} {\frac{{\log (1 + k)}}{k}} = \int_1^{n - 1} {\frac{{\log (1 + t)}}{t}{\rm d}t} & - 2\int_0^{ + \infty } {\frac{1}{{{\rm e}^{2\pi y} - 1}}{\mathop{\rm Im}\nolimits}\! \left( {\frac{{\log (2 + {\rm i}y)}}{{1 + {\rm i}y}}} \right){\rm d}y} \\ & + \frac{{\log 2}}{2} + \mathcal{O}\!\left( {\frac{{\log n}}{n}} \right). \end{align*} Here $$ - 2\int_0^{ + \infty } {\frac{1}{{{\rm e}^{2\pi y} - 1}}{\mathop{\rm Im}\nolimits} \!\left( {\frac{{\log (2 + {\rm i}y)}}{{1 + {\rm i}y}}} \right){\rm d}y} = \int_0^{ + \infty } {\frac{1}{{{\rm e}^{2\pi y} - 1}}\frac{{y\log (4 + y^2 ) - 2\arctan (y/2)}}{{y^2 + 1}}{\rm d}y} $$ and $$ \int_1^{n - 1} {\frac{{\log (1 + t)}}{t}{\rm d}t} = \frac{\log ^2 n}{2} + \frac{{\pi ^2 }}{{12}} + \mathcal{O}\!\left( {\frac{{\log n}}{n}} \right). $$ Accordingly, $$ \prod\limits_{k = 2}^n {\sqrt[{k - 1}]{k}} = Cn^{\frac{1}{2}\log n} \left( {1 + \mathcal{O}\!\left( {\frac{{\log n}}{n}} \right)} \right) $$ as $n\to +\infty$, where $$ \log C = \frac{{\pi ^2 }}{{12}} + \frac{{\log 2}}{2} + \int_0^{ + \infty } {\frac{1}{{{\rm e}^{2\pi y} - 1}}\frac{{y\log (4 + y^2 ) - 2\arctan (y/2)}}{{y^2 + 1}}{\rm d}y} = 1.18493104146 \ldots $$ Addendum. Another way is to note that $$ \sum\limits_{k = 1}^{n - 1} {\frac{{\log (1 + k)}}{k}} = \sum\limits_{k = 1}^{n - 1} {\frac{{\log k}}{k}} + \sum\limits_{k = 1}^{n - 1} {\frac{{\log (1 + 1/k)}}{k}} = \frac{{\log ^2 n}}{2} + \gamma _1 + \sum\limits_{k = 1}^\infty {\frac{{\log (1 + 1/k)}}{k}} + o(1) $$ where $\gamma_1$ is one of the Stieltjes constants. For the infinite series see this answer of mine or $\text{A}131688$ in the OEIS.

Answer 2

Let's denote $$f(n) = \prod_{k=2}^{n} \sqrt[k-1]{k}=e^{g(n)}$$ where $$g(n)=\sum_{k=1}^{n-1}\frac{\ln(1+k)}k$$ To find the asymptotic $g(n)$ at $n\to\infty$ we can use the Euler-Maclaurin' formula

In our case $g(x)=\frac{\ln(1+x)}x$. Putting into the formula and taking derivatives, $$g(n)=\sum_{k=1}^{n-1}\frac{\ln(1+k)}k\sim\int_1^{n-1}\frac{\ln(1+x)}xdx+\frac12\Big(\ln2+\frac{\ln n}{n-1}\Big)+\frac1{12}\left(\frac1{n(n-1)}-\frac{\ln n}{(n-1)^2}-\frac12+\ln2\right)-\frac1{30}\frac1{4!}\left(...\right)+......\tag{1}$$ To evaluate the integral, we make the substitution $t=\frac1x$ $$\int_1^{n-1}\frac{\ln(1+x)}xdx=\int_{1/(n-1)}^1\frac{\ln(1+t)}tdt-\int_{1/(n-1)}^1\frac{\ln t}tdt$$ $$=\frac12\ln^2n+\frac{\pi^2}{12}+O\left(\frac{\ln n}n\right)\tag{2}$$ Putting (2) into (1) and dropping vanishing at $n\to\infty$ terms, $$g(n)\sim\frac{\ln^2n}2+\frac{\pi^2}{12}+\frac{\ln 2}2+\frac{\ln 2}{12}-\frac1{24}-\frac{\ln 2}{120}+\frac1{720}+...$$ We accurately evaluated the growing term, but only approximated the constant term (though, approximated with a high accuracy). I'm not sure whether we can get a constant term in a closed form in this case. $$f(n)=e^{g(n)}\sim e^{\frac{\pi^2}{12}+\frac{\ln 2}2+\frac{\ln 2}{12}-\frac1{24}-\frac{\ln 2}{120}+\frac1{720}+...}\,e^{\frac12\ln^2n}$$ $$\boxed{\,\,f(n)\approx e^{\frac{\pi^2}{12}+\frac{23}{40}\ln2-\frac{29}{720}}\,\sqrt{n^{\ln n}}\,\,}$$

---

Numeric check with WolframAlpha

$\displaystyle n=100 \qquad \text{exact}\, = 127 471\qquad \text{approximation}\, =131 209$ $\displaystyle n=1000 \,\,\quad \text{exact}\, = 7.48699 \,\cdot 10^{10}\quad \text{approximation}\, =7.48903 \,\cdot 10^{10}$

Answer 3

Using the same approach as @Svyatoslav

$$\color{blue}{S_n=\sum_{k=2}^n \frac{\log (k)}{k-1}= C +\frac{1}{2}\log ^2(n)- \frac 1{2n} P(n) \log(n)- \frac 1{n} Q(n)}$$ where $$P(n)=1+\frac{1}{6 n}-\frac{1}{60 n^3}+\frac{1}{126 n^5}-\frac{1}{120 n^7}+\frac{1}{66 n^9}+\cdots$$ $$Q(n)=1+\frac{1}{6 n}+\frac{1}{36 n^2}-\frac{1}{180 n^3}-\frac{13}{1800 n^4}+\frac{1}{1512 n^5}+\frac{47}{8820 n^6}+\frac{1}{1575 n^7}-\frac{1703}{226800 n^8}+\cdots$$ For this level of expansion $$C=\frac{\pi ^2}{12}+\frac{8268803161823011 }{160626866400}\log (2)-\frac{4331418924726329249667169301}{121390492708302225408000}$$

Some results with $n=10^m$ $$\left( \begin{array}{ccc} m & \text{approximation} & \text{exact} \\ 1 & 3.617187748 & 3.617140113 \\ 2 & 11.75569398 & 11.75564634 \\ 3 & 25.03906555 & 25.03901792 \\ 4 & 43.59960303 & 43.59955540 \\ 5 & 67.45863749 & 67.45858986 \\ 6 & 96.61913676 & 96.61908912 \\ 7 & 131.0814815 & 131.0814338 \\ \end{array} \right)$$

Edit

After @Gary's solutions and comments, I propose $$C=\frac{{\pi ^2 }}{{12}} + \frac{{\log 2}}{2}+\frac{3559-5 \sqrt{487001}}{4388}$$ since $$\frac{3559-5 \sqrt{487001}}{4388} =\color{red}{0.0158904177566}734\cdots$$ while $$\int_0^\infty \frac{1}{e^{2 \pi y}-1} \frac{y \log \left(y^2+4\right)-2 \tan ^{-1}\left(\frac{y}{2}\right)}{y^2+1}\,dy=0.0158904177566070$$