The rth term in $(1+x)^{1/x} = e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ Let 
$$(1+x)^{\frac {1}{x}} = e.G(x)$$
Taking logarithm on both sides,
$$\frac {1}{x} \log {(1+x)} = 1 + \log {G(x)}$$
Putting in the Taylor expansion for $\log {(1+x)}$ we have,
$$\frac {1}{x}(x - \frac {1}{2}x^2 + \frac {1}{3}x^3 - ....) = 1 + \log {G(x)}$$
Solving for $G(x)$ we have,
$$G(x) = e^{-\frac {1}{2}x + \frac {1}{3}x^2 - ...}$$
The difficulty starts here, for in order to get the desired Taylor expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ I have to plug in the entire expansion $-\frac {1}{2}x + \frac {1}{3}x^2 - ...$ for the variable in the Taylor expansion for $e$. Is there any alternative way? My main question is : what is the rth term of the desired expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ in its closed form?
 A: This is a slight rewriting of a problem solved in Math Overflow 77389.  That answer can be put in the form
$$ (1+x)^{1/x} = e\,\sum_{n=0}^\infty a_n x^n \text{ with } a_n=\sum_{k=0}^n \frac{S_1(n+k, k)}{(n+k)!} \sum_{m=0}^{n-k}\frac{(-1)^m}{m!} $$
$S_1(n,k)$ is the Stirling number of the first kind, StirlingS1[n,k], in Mathematica.
A: Let $G(x)=(1+x)^{1/x}$
$\ln G(x)=\dfrac{\ln(1+x)}x=1-\dfrac x2+\dfrac{x^2}3-\dfrac{x^3}4+\cdots$
$$G(x)=e\cdot e^{-x/2}\cdot e^{-x^2/3}\cdot e^{-x^3/4}\cdots$$ ignoring terms containing $x^4$
$$=e\left(1-\dfrac{x}2+\dfrac{\left(-\dfrac x2\right)^2}{2!}+\dfrac{\left(-\dfrac x3\right)^3}{3!}+\cdots\right)\left(1-\dfrac{x^2}3+\dfrac{\left(-\dfrac{x^2}3\right)^2}{2!}+\cdots\right)\left(1-\dfrac{x^3}4+\cdots\right)\cdots$$
$$=?$$
A: Let $$f(x)=\sum^\infty_{k=1} \frac{(-1)^k}{k+1}x^k$$
Then, $G(x)=e^f$.
To expand $G$ in Taylor series up to the cube term, we have to compute up to the third derivative of $G$.
This isn’t too difficult.
$$G(0)=1$$
$$G’(0)=f’(0)=-\frac12$$
$$G’’(0)=f’’(0)+f’^2(0)=2\cdot \frac13+\frac14$$
$$G’’’(0)=f’’’(0)+2f’(0)f’’(0)+f’’(0)f’(0)+f’^3(0)$$
A: \begin{align}
\dfrac1e(1+x)^{1/x} 
&= \exp\left(\dfrac{\ln(1+x)-x}{x}\right) \\
&= \exp\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right) \\
&= 1+\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right)+\dfrac12\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right)^2 \\
& +\dfrac16\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right)^3+\cdots\\
&= 1-\frac{1}{2}x+\left(\dfrac13+\dfrac18\right)x^2+\left(-\dfrac14+\dfrac122\dfrac{-1}{2}\dfrac13+\dfrac16\dfrac{-1}{8}\right)x^3+\cdots
\end{align}
