1

This question is a direct continuation of my previous question.

How to prove the formula for the Reciprocal Multifactorial constant?

User metamorphy provided a lucid proof for the closed form equation of the Reciprocal Multifactorial Constant. And also a statement on its asymptote as k approaches infinity. Which is as follows:

$m(k)=1+e^{1/k}(H_k-\Delta_k)\ \text{and,}\\ > \Delta_k=\frac1k\underbrace{\int_0^1\frac{1-t}{1-t^{1/k}}t^{1/k}\frac{1-e^{-t/k}}{t}\,dt}_{\text{has > a finite $k\to\infty$ limit}}\underset{k\to\infty}{\longrightarrow}0.$

I was aware that the Reciprocal Multifactorial would eventually approach a harmonic series due to the definition of a multifactorial, which is as follows:

$n!^{(k)} = \begin{cases} 1 & \text{if $n=0$} \\ n & \text{if $0<n\leq k$} \\ n\left((n-k)!^{(k)}\right) & \text{if $n>k$} \end{cases}$

As k approaches infinity the $k^{th}$ multifactorial will tend towards the natural numbers. And the reciprocal series would approach $1+H_k$

However I am asking this question to know how to prove this using the closed form formula as metamorphy did in the previous question. I wasn't able to solve it myself.

Unrelated: Do you think there are any other interesting properties about this series worth examining, since I've found it quite interesting?

Bhoris Dhanjal
  • 407
  • Your actual question is not clear to me. Are you asking how metamorphy is arguing $\Delta_k \to 0$? It looks like the integrand is strictly increasing on $[0, 1]$, so the maximum is at $t=1$. In that case, the $k \to \infty$ limit works out to be $k(1-e^{-1/k})$, which itself has limit $1$ as $k \to \infty$. So, the integrand is basically bounded by $1$. You can make this more precise, or find your own analysis of the integrand, if that's what you're interested in. – Joshua P. Swanson Feb 11 '21 at 10:06
  • I understood why $\Delta_k$ approaches zero. I meant how do I derive $m(k)=1+e^{1/k}(H_k-\Delta_k)$ for k approaches infinity from the general form of $m(k)=1+\frac{e^{1/k}}k\sum_{r=1}^k k^{r/k}\int_0^{1/k}x^{r/k-1}e^{-x},dx$. Hope this clarifies it. – Bhoris Dhanjal Feb 11 '21 at 10:16

1 Answers1

1

I think I have to answer it myself. Let the point of departure be $$m(k)=1+\frac{e^{1/k}}{k}\sum_{r=1}^k\int_0^1 t^{r/k-1}e^{-t/k}\,dt,$$ appearing in my post immediately before the substitution $t=kx$.

The statement being discussed just comes from $e^{-t/k}=1-(1-e^{-t/k})$: $$\int_0^1 t^{r/k-1}e^{-t/k}\,dt=\frac{k}{r}-\int_0^1 t^{r/k-1}(1-e^{-t/k})\,dt,$$ giving $m(k)=1+e^{1/k}(H_k-\Delta_k)$ as written: $$\Delta_k=\frac1k\int_0^1\left(\sum_{r=1}^k t^{r/k}\right)\frac{1-e^{-t/k}}{t}\,dt=\frac1k\int_0^1\frac{(1-t)(1-e^{-t/k})}{t(t^{-1/k}-1)}\,dt.$$

For curiosity, here's the asymptotics of $\Delta_k$ in more detail:

\begin{align*} \Delta_k&=\frac1k f\Big(\frac1k\Big),\\ f(\lambda)&=\int_0^1\frac{(1-t)(1-e^{-\lambda t})}{t(e^{-\lambda\log t}-1)}\,dt\asymp\sum_{n=0}^{(\infty)}a_n\lambda^n,\\ a_0&=\int_0^1\frac{1-t}{-\log t}\,dt=\color{blue}{\log 2},\\ a_1&=-\frac12\int_0^1\frac{1-t}{-\log t}(t-\log t)\,dt=\color{blue}{-\frac14-\frac12\log\frac32},\\ a_2&=\frac{1}{12}\int_0^1\frac{1-t}{-\log t}(2t^2-3t\log t+\log^2 t)\,dt=\color{blue}{\frac{5}{48}+\frac16\log\frac43},\\ a_3&=-\frac{1}{24}\int_0^1\frac{1-t}{-\log t}(t^3-2t^2\log t+t\log^2 t)\,dt=\color{blue}{-\frac{11}{864}-\frac{1}{24}\log\frac54}, \end{align*} and so on. For $m(k)$, use the power series of $e^{1/k}$, and known asymptotics of $H_k$...

metamorphy
  • 43,591