Is this a known factorial approximation?

Question

While playing with the fixed points (i.e. $e^{\pm\pi/3}$) of the iterated composition of $(1-x)^{-1}$ and a kind of unitary transform, I stumbled across what I believed to be an identity that connects the factorial and the geometric series.

For $n=1$, it converges to $(1-e)^{-1}$ and always seems to point the nearest integer as the proper value of $n!$ $$ n! \approx \frac{-1}{1-\sum_{k=0}^{\infty} \frac{1}{(n*k)!}} |n \in Z^{+} $$ So my question is twofold, is it a known identity? And how does one manage to prove this since the sum diverges at $n=0$ and $\forall\ {}n \in Z, n!\ge1$. It seems to make it a difficult task.

Answer 1

The "identity" is an approximation; playing with your Wolfram Alpha link shows that the approximation gets better and better as $n$ gets larger, but is very inexact for small $n$.

And, as Karl noted in the comments, the approximation is equivalent to $\sum_{k=2}^\infty \frac{1}{(nk)!} \overset{?}{\approx} 0$, which does get better as $n$ is larger.

Answer 2

This approximation is actually a degenerated tautology. the tautology is the partial sum up to k=1. And the approximation worsen the more k terms are added.

$$ n! = \frac{-1}{1-\sum_{k=0}^{k=1} \frac{1}{(n·k)!}} $$ $$ n! = \frac{-1}{1-(1+1/n!)} $$ As Karl pointed out, the series starting from k=2 converges to 0 as n get bigger. $$ lim_{n\rightarrow\infty} \sum_{k=2}^{\infty}\frac{1}{(n·k!)}=0 $$

This is very interesting because it emerged as the result of a kind of unitary transform. Maybe it has some interest in algebra because it can be seen as an approximation for $$ n! = \frac{-1}{1-(1+\sum_{k=0}^{\infty}\frac{1}{n!\delta_{nk}})} $$ Which does not translate into a geometric series because it diverges to infinity $$ \sum_{j=0}^{\infty} (1+\sum_{k=0}^{\infty}\frac{1}{n!\delta_{nk}})^{j} \rightarrow \infty $$

Answer 3

For the fun of playing with huge numbers, let $$a_n=\log_{10}\Bigg[n! \left(1-\sum _{k=0}^{\infty } \frac{1}{(k n)!}\right)+1\Bigg]$$ A quick and dirty nonlinear regression of the totally empirical model $$a_n= \alpha - \beta \, n^\gamma$$ gives, with $R^2=0.9999965$

$$\begin{array}{llll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \alpha & 0.429416 & 0.019544 & \{0.379177,0.479655\} \\ \beta & 0.588406 & 0.008644 & \{0.566185,0.610627\} \\ \gamma & 1.319139 & 0.006204 & \{1.303191,1.335087\} \\ \end{array}$$

More than likely, looking at the value of $\gamma$, we can suspect a logarithmic contribution.

Edit

If you want nicer, define $$b_n=\sum _{k=0}^{\infty } \frac{1}{(k n)!}$$ For $1 \leq n \leq 4$, they have simple closed forms $$\left\{e,\cosh (1),\frac{e^{3/2}+2 \cos \left(\frac{\sqrt{3}}{2}\right)}{3 \sqrt{e}},\frac{1}{2} (\cos (1)+\cosh (1))\right\}$$ and the next are $$b_5=\, _0F_4\left(;\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{5^5}\right)$$ $$b_6=\, _0F_5\left(;\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};\frac{1} {6^6}\right)$$ $$b_7=\, _0F_6\left(;\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7};\frac{1}{7^7}\right)$$ and the pattern is clear $$b_n=\, _0F_{n-1}\left(;\frac{1}{n},\frac{2}{n},\cdots,\frac{n-1}{n};\frac{1}{n^n}\right)$$

and with $R^2=0.99999865$ the model $$\log(\log(b_n))=\alpha - \beta \, n^\gamma$$ $$\begin{array}{llll} \text{} & \text{Estimate} & \text{Std. Error} & \text{Confidence Interval} \\ \alpha & 1.606083 & 0.068886 & \{1.452595,1.759571\} \\ \beta & 0.669080 & 0.010914 & \{0.644762,0.693397\} \\ \gamma & 1.397903 & 0.005155 & \{1.386417,1.409389\} \\ \end{array}$$