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Was curious about this after seeing that $$ \lim_{ x \rightarrow 0}{ \left[ \cfrac{1}{x} \cdot \left((1+x)^{1/x} - e\right) \right]} = -e/2 $$ An answer to my question would solve the analogous questions about the higher-order rates of convergence of $ (1+x)^{1/x} $ to $ e $. But it is hard to find order in the derivatives of $ (1+x)^{1/x} $. Wondering if this sequence has been studied. Here's a little computation: Mathematica computation of first 9 Taylor coefficients

Edit: it is interesting to observe these values when they are converted to decimal form. They are approaching $(-1)^n$. Viz., enter image description here A proof of this fact would be interesting if a closed-form formula cannot be found.

Jake Mirra
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    By the way, in Mathematica D[expr,{x,n}] is the $n$th derivative of expr with respect to $x$. There's no need to define your own function as you seem to have done. – YiFan Tey Oct 13 '19 at 20:50

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As per https://oeis.org/A055505 and https://oeis.org/A055535, the coefficient can be expressed as a finite sum in terms of Stirling numbers of the first kind ${n\brack k}$, so it would be $$ f(x)=\sum_{n\geq 0}\left(e\sum_{k=0}^{n}{n+k\brack k}\frac1{(n+k)!}\sum_{l=0}^{n-k}\frac{(-1)^l}{l!}\right)x^n, $$ or when expressing the inner sum in terms of Incomplete gamma function $$ f(x)=\sum_{n\geq 0}\left(\sum_{k=0}^{n}{n+k\brack k}\frac{\Gamma(n-k+1,-1)}{(n+k)!(n-k)!}\right)x^n. $$

Now I don't know the derivation behind this, but by Maple verification it seems to be correct (someone else can surely give more details on math behind this).

Sil
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  • Accepted this answer. I still haven't done the exercise of showing that these coefficients approach $ \pm 1 $ as $ n \rightarrow \infty $, but this is satisfying enough for me. – Jake Mirra Oct 17 '19 at 02:57