Is there a series representation for $\frac{1}{\log(x)}$?

Question

for $x>0$

it is known that: $$\log(x)=2\sum_{k=1}^\infty \frac{\frac{x-1}{x+1}^{2k-1}}{2k-1}$$ Is there a series representation for $\frac{1}{\log(x)}$ in the following form? $$\frac{1}{\log(x)}=a_1(x)+a_2(x)+a_3(x)+\dots=\sum_{k=1}^\infty a_k(x)$$ Or, is there a way to transform the expression $$ \frac{1}{2\sum_{k=1}^\infty \frac{\frac{x-1}{x+1}^{2k-1}}{2k-1}}$$ into the aforementioned form?

Answer 1

Developed as a truncated series around $x=a >1$, you have $$\frac{L}{\log (x)}=1-t+\left(\frac{L}{2}+1\right) t^2-\left(\frac{L^2}{3}+L+1\right) t^3+\left(\frac{L^3}{4}+\frac{11 L^2}{12}+\frac{3 L}{2}+1\right) t^4-\left(\frac{L^4}{5}+\frac{5 L^3}{6}+\frac{7 L^2}{4}+2 L+1\right) t^5+\left(\frac{L^5}{6}+\frac{137 L^4}{180}+\frac{15 L^3}{8}+\frac{17 L^2}{6}+\frac{5 L}{2}+1\right) t^6-\left(\frac{L^6}{7}+\frac{7 L^5}{10}+\frac{29 L^4}{15}+\frac{7 L^3}{2}+\frac{25 L^2}{6}+3 L+1\right) t^7+\left(\frac{L^7}{8}+\frac{363 L^6}{560}+\frac{469 L^5}{240}+\frac{967 L^4}{240}+\frac{35 L^3}{6}+\frac{23 L^2}{4}+\frac{7 L}{2}+1\right) t^8+O\left(t^{9}\right)$$ where $L=\log(a)$ and $t=\frac{x-a}{a L}$.

If $a=e$, we have the nice $$\frac{1}{\log (x)}=1-t+\frac{3 }{2}t^2-\frac{7 }{3}t^3+\frac{11 }{3}t^4-\frac{347 }{60}t^5+\frac{3289 }{360}t^6-\frac{1011 }{70}t^7+\frac{38371 }{1680}t^8+O\left(t^{9}\right) \qquad \text{where}\qquad t=\frac{x-e}{e}$$

Edit

For the case where $a=e$, writing $$\frac{1}{\log (x)}=\sum_{n=0}^\infty (-1)^n\, \Big[ \sum _{k=0}^n k! \left|S_n^{(k)}\right|\Big]\frac{t^n} {n!}$$ found using sequence $A007840$ in $OEIS$.

Answer 2

Your premise is faulty. $f$ having a series centered at $a$ does not mean $\frac{1}{f}$ can have a series centered at $a$, too. However, if it is points far to the right you care about, we can find a series for $\frac{1}{\log x}$ about powers of $e$. Take for $e$, for example:

$$f(e) = 1$$ $$f'(e) = -\frac{1}{e}$$ $$f''(e) = \frac{3}{e^2}$$

to get

$$\frac{1}{\log x} = 1 - \left(\frac{x}{e}-1\right) + \frac{3}{2}\left(\frac{x}{e}-1\right)^2 + \cdots$$

and you can use log properties to center this around the power of $e$ closest to your value of choice.

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$\mathbf{EDIT}:$ Given your expression for a series $\sum_n a_n(x)$ is just a sum of functions and not necessarily a power series, we can have the following thing:

$$\frac{1}{\log(x)} = \frac{1}{n+\log\left(\frac{x}{e^n}\right)}$$

where $n$ is chosen such that $x<e^{2n}$. Then we have that

$$\frac{1}{\log x} = \frac{1}{n}\sum_{k=0}^\infty \frac{(-1)^k\log^k \left(\frac{x}{e^n}\right)}{n^k}$$