<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 $$f(x)=\sum^\infty_{k=1} \frac{(-1)^k}{k+1}x^k$$
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.
$$G(0)=1$$
$$G’(0)=f’(0)=-\frac12$$
$$G’’(0)=f’’(0)+f’^2(0)=2\cdot \frac13+\frac14$$
$$G’’’(0)=f’’’(0)+2f’(0)f’’(0)+f’’(0)f’(0)+f’^3(0)$$
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