Find the Taylor series expansion of the following function around $x=0$
$$f(x)=\frac{1}{\ln(1+\frac{1}{x})+1}$$
My solution:
$$f(0)=\lim_{x \to 0}f(x)=0$$ $$f'(0)=\lim_{x\to 0}f'(x)=\lim_{x\to 0}\frac{1}{(x^2 +x) (\ln(1+\frac{1}{x})+1)^2}=0$$ $$f''(0)=\lim_{x\to 0}f''(x)=\lim_{x\to 0}\frac{2}{(\ln(1+\frac{1}{x})+1)^3x^4(1+1/x)^2}-\frac{2}{(\ln(1+1/x)+1)^2 x^3 (1+1/x)}+\frac{1}{(\ln(1+1/x)+1)^2 x^4 (1+1/x)^2}=-\infty$$
It looks like something is wrong, What should I do now?
MAPLE gives me the following. It does not work for me since the terms behind $x^{K}$ depend on $x$:
$$\frac{a}{\ln(1/x)+a}-\frac{a x}{(\ln(1/x)+a)^2}+\frac{(1/2)a/(\ln(1/x)+a)+a/(\ln(1/x)+a)^2}{\ln(1/x)+a}x^2$$
Because $\ln x$ has a non-removable singularity at $x=0$, your expression would not be expected to have a valid Taylor series (or even a valid MacLauren series starting from some negative power of $x$) about $x=0$.
The Taylor series about $x=1$ starts with $$ \frac{1}{1+\ln 2}+\frac1{2(1+\ln 2)^2}(x-1)-\frac{3\ln 2 + 1}{8(1+\ln 2)^3}(x-1)^2\cdots $$