<s>[INST] 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?
 [/INST] 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 $x^4$
$$=e\left(1-\dfrac{x}2+\dfrac{\left(-\dfrac x2\right)^2}{2!}+\dfrac{\left(-\dfrac x3\right)^3}{3!}+\cdots\right)\left(1-\dfrac{x^2}3+\dfrac{\left(-\dfrac{x^2}3\right)^2}{2!}+\cdots\right)\left(1-\dfrac{x^3}4+\cdots\right)\cdots$$
$$=?$$
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