MHB How can I develop ln(x) into a series for x >= 1 in fluid dynamics?

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To develop ln(x) into a series for x ≥ 1 in fluid dynamics, one approach is to use the series for ln(1+x) and ln(1-x). By writing the series for ln(1+x) and ln(1-x), and then subtracting the latter from the former, you can derive the series for ln((1+x)/(1-x)). This method is particularly useful since x represents a non-dimensional radial coordinate in an annular tube, where x starts at 1 and increases. The derived series can help in analyzing fluid behavior in such geometries. This approach provides a valid solution for the specified condition of x ≥ 1.
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I need to develop $\mathrm{ln}(x)$ into series, where $x \geq 1$, and I don`t know how? In literature I only found series of $\mathrm{ln}(x)$, where:

1. $|x-1| \leq 1 \land x \neq 0$, $ \,\,\,\,\, \mathrm{ln}(x) = x - 1 - \dfrac{(x-1)^2}{2} + ...$ 2. $|x| \leq 1 \land x \neq -1$, $ \,\,\,\,\, \mathrm{ln}(x+1) = x - \dfrac{x^2}{2}+ ...$

My problem is problem in area of fluid dynamics, and $x$ is non-dimensional coordinate and it signifies radial coordinate of annular tube (it starts in the center of the tube). At the wall of inner tube $x=1$, and at the wall of outer tube it only can be larger (and values are not limited), because of that I need to fulfill a condition $x \geq 1$, for developing $\mathrm{ln}(x)$ into series.
 
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Write series for $$\ln (1+x)$$ and $$\ln (1-x)$$. Then subtract second from the first one and you'll obtain the series for $$\ln \left( \frac{1+x}{1-x} \right) $$, which gives you one possible series.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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