Proof about the power series of reciprocal multifactorials $m_x(k)=\sum_{n=0}^\infty \frac{x^n}{n\underbrace{!\cdots!}_{\text{k times}}}$

Question

The proof I've attempted mimics very closely the answer on this question.

How to prove the formula for the Reciprocal Multifactorial constant?

Pre-requisite definitions:

• A multifactorial of order $k \in \mathbb{N}$ is defined for $n \in \mathbb{N}_0$ as $n!\underbrace{\cdots!}_\text{k times}=n(n-k)(n-2k)\cdots$
  and $0!\underbrace{\cdots!}_\text{k times}=1$
• $\text{B}(x, y)=\int_0^1 t^{x-1}(1-t)^{y-1} dt$
• $\gamma (a, x) = \int_0^x t^{a-1} e^{-t}$

Here is what I have done,

$\begin{align*}
m_x(k)&=\sum_{n=0}^{\infty}\frac{x^n}{n!_{(k)}}=1+\sum_{r=1}^{k}\sum_{q=0}^{\infty}\frac{x^{kq+r}}{(kq+r)\underbrace{!\dots!}_{\text{k times}}}\\
&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{x^{kq+r}}{k^{q+1}q!}\mathrm{B}\left(\frac rk,q+1\right)\\
&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{x^{kq+r}}{k^{q+1}q!}\int_0^1 t^{r/k-1}(1-t)^q\,dt\\
&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{x^{kq}x^r}{k k^{q}q!}\int_0^1 t^{r/k-1}(1-t)^q\,dt\\
&=1+\frac1k\sum_{r=1}^k\int_0^1 x^r t^{r/k-1}\sum_{q=0}^\infty\frac{1}{q!}\left(x^k\frac{1-t}{k}\right)^q dt\\
&=1+\frac1k\sum_{r=1}^k\int_0^1 x^r t^{r/k-1}e^{(x^k-x^k t)/k}\,dt\\
&\text{Now we substitute, }t=kx^{1-k}, dt=(1 – k) k x^{-k}\\
&=1+\frac1k\sum_{r=1}^k\int_0^{x^k/k} x^r (kx^{1-k})^{r/k-1}e^{(x^k-x^k kx^{1-k})/k}((1 – k) k x^{-k})\
dx\\
&=1+\frac{e^{x^k/k}}{k}\sum_{r=1}^k\int_0^{x^k/k} x^r (kx^{1-k})^{r/k-1}e^{(-x^k kx^{1-k})/k}((1 – k) k x^{-k})\
dx\\
&=1+\frac{e^{x^k/k}}{k}\sum_{r=1}^k\int_0^{x^k/k} (1-k) (e^{-x}) k^{r/k} x^{r/k – 1}\ dx\\
&=1+\frac{e^{x^k/k}}{k}\sum_{r=1}^k (1-k)k^{r/k} \int_0^{x^k/k} x^{r/k – 1} e^{-x}\ dx\\
&=1+\frac{e^{x^k/k}}{k}\sum_{r=1}^k (1-k)k^{r/k} \gamma \left (\frac{r}{k}, \frac{x^k}{k} \right)\\
\end{align*}$

However, this final answer appears to be completely wrong. I'm not sure if there's just some bad arithmetic mistake.

I have tried verifying it using some computations in Mathematica (by comparing it with the summation definition described in the first line of the proof). But the two answers are way off.

I think the actual answer must be of the following form,

$$m_x(k)=1+\frac{e^{x^k/k}}{k}\sum_{r=1}^{k}k^{r/k}\gamma\left ( \frac{r}{k}, \frac{x^k}{k} \right )$$

Computed values for this expression perfectly coincide with the values computed with the summation definition.

Thanks for any and all help!

PS: I have already read a similar question Evaluating the power series of the multifactorial reciprocal $\sum_{n=0}^\infty \frac{x^n}{n!!…!!}$.
But the method of proof is completely different.

Answer 1

Using $x$ in the substitution is totally wrong, since $x$ is effectively a constant in this case.

The correct way to proceed is to factor out $x^r$ and $e^{x^k}$ from the integral, followed by substituting $u=x^{k} t/k$. Everything then falls into place.