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Over the last few weeks I have become obsessed with expressions like $$ \frac{e+4 e^{2}+e^{3}}{(1-e)^{4}}, $$ $$ \frac{e+26 e^{2}+66 e^{3}+26 e^{4}+e^{5}}{(1-e)^{6}}, $$ or $$ \frac{e+120 e^{2}+1191 e^{3}+2416 e^{4}+1191 e^{5}+120 e^{6}+e^{7}}{(1-e)^{8}}. $$

These expressions sparked interest in me because they approximate $3!,5!$ and $7!$, respectively and are part of a larger family of decent approximations (these approximations are quite good till 16!).

One reason why this is true is because the relation between this expressions and polylogarithm, and the later with the gamma function, but I would really like to know if there is another way to justify why they approach factorials.

I have delved on analytic combinatorics, and $q$-analogs that talk about evaluating at roots of unity, poles or saddle points, but none of them seem totally appropriate, Although they study the idea of evaluating a generating function in transcendental or complex numbers, none of them seem to relate directly to this.

PD: a Dual to this identities is the evaluation at $e^{-1}$, $$\frac{e^{-1}+11 e^{-2}+11 e^{-3} +e^{-4 }}{\left(1-e^{-1}\right)^{5}}$$ these are the same but with a correcting minus sign, so with $e^{-1}$ we get $4!$ while with $e$ we get $-4!$. I add these as they make more sense when we convert back to the classic polylogarithms expression $$5! \approx \operatorname{Li}_{-5}(z) = \sum_{k=0}^ \infty k^{5}z^k$$ with $z=\frac{1}{e}$ that needs $|z|<1$, but I guess the $e$ expressions coincide with the analytic continuation of these series

Also Something that made me stick with this subject was that for example $$ \frac{z+26 z^{2}+66 z^{3}+26 z^{4}+z^{5}}{(1-z)^{6}} =\frac{1}{z-1}+\frac{31}{(z-1)^{2}}+\frac{180}{(z-1)^{3}}+\frac{390}{(z-1)^{4}}+\frac{360}{(z-1)^{5}}+\frac{120}{(z-1)^{6}} $$ where something notable is that the last coefficient is exactly 5!, this is because eulerian numbers sum to factorials, but this also seemed crazily connected with the Cauchy residue theorem, only that this one was about $(z-1)^{-n}$ and not about $(z-1)^{-1}$

EDIT: Someone edited out an oeis entry linking this to the eulerian polynomials, so just so you know I'm aware what they are, also I will add a comment that I think it's crucial to finding the connection

"This seems to be connected to this combinatorial problem math.stackexchange.com/questions/257890/simon-newcombs-problem stating the formula $$\sum_{d=0}^{\infty} \frac{A_{d}(t) x^{d}}{(1-t)^{d+1} d !}=\frac{1}{1-t e^{x}}$$ the wikipedia page has the case $x=1$ although it doesn't seem to provide any source, while Jair answers [Comment] is that formula with $t=1/e$. This seems promising as, that formula is almost a composition of classical analytical combinatorics constructions, and the wikipedia case has a $1/e$ as pole. "

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    Connected: https://math.stackexchange.com/q/1763379 https://artofproblemsolving.com/community/c7h84775p492710 found using the formula searching tool https://approach0.xyz/ . – Jean Marie Jun 16 '21 at 16:48
  • Yes I know the connection to the series and derivates, but I would like another approach connected to analytic combinatorics. – Alejandro Quinche Jun 16 '21 at 17:27
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    Not an answer, but take a look at the Taylor series for $\frac{1}{1-e^{x-1}}$: https://www.wolframalpha.com/input/?i=taylor+series+of+1+%2F+%281-e%5E%28x-1%29%29 – Jair Taylor Jun 16 '21 at 17:41
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    The coefficients of that series are your sums (divided by factorials.) As $x \rightarrow 0$ we have $\frac{1}{1-e^{x-1}} \approx 1 / (1 - (1 + x-1) = 1/(1-x) = 1 + x + x^2 + \cdots$ and equating the coefficients then gives your approximations. This is not rigorous though. – Jair Taylor Jun 16 '21 at 17:44
  • (This is neat btw.) – Jair Taylor Jun 16 '21 at 17:45
  • This seems to be connected to this combinatorial problem https://math.stackexchange.com/questions/257890/simon-newcombs-problem stating the formula $\sum_{d=0}^{\infty} \frac{A_{d}(t) x^{d}}{(1-t)^{d+1} d !}=\frac{1}{1-t e^{x}}$ the wikipedia page has the case $x=1$ although it doesn't seem to provide any source, while Jair answers is that formula with $t=1/e$. This seems promising as, that formula is almost a composition of classical analytical combinatorics constructions, and the wikipedia case has a $1/e$ as pole. Any hints on how you came up with that formula? – Alejandro Quinche Jun 17 '21 at 17:22

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