Proof for the series $\ln{x} = \sum_{k=1}^\infty\frac1k\left(\frac{x-1}{x}\right)^k,\ \forall x>\frac12 $ The equation only came from my observation via brute force, graph, and Wolfram|Alpha, but I do not know where it exactly came from, except that I noticed in my attempts to prove this that it resembles the Taylor Series of $\ln(x)$ centered at $1$.
$$\ln{x} = \sum_{k=1}^\infty\frac1k\left(\frac{x-1}{x}\right)^k,\ \forall x>\frac12 $$
Thank you!
 A: Let $$f(y)=\sum_{k=1}^\infty \frac{y^k}k,\ |y|<1$$  Then $f(0) = 0$ and $$f'(y)=\sum_{k=1}^\infty y^{k-1}=\frac1{1-y}$$  so that $$f(y)=\int_0^y\frac{\mathrm{d}x}{1-x}=-\ln(1-y)$$  Now if $\ln{x}=-\ln(1-y)$, then $x=\frac1{1-y}$ or $y=\frac{x-1}{x}$.  This gives your formula, vaild when $$\left|\frac{x-1}{x}\right|<1,$$ that is, when $x>\frac12$. 
A: Let $ n $ ne a positive integer and $ x $ a real such that $ \left|x\right|<1 \cdot $
\begin{aligned} \sum_{k=1}^{n}{\frac{x^{k}}{k}}+\ln{\left(1-x\right)}&=\int_{0}^{x}{\sum_{k=1}^{n}{t^{k-1}}\,\mathrm{d}t}-\int_{0}^{x}{\frac{\mathrm{d}t}{1-t}}\\ &=\int_{0}^{x}{\frac{1-t^{n}}{1-t}\,\mathrm{d}t}-\int_{0}^{x}{\frac{\mathrm{d}t}{1-t}}\\ \sum_{k=1}^{n}{\frac{x^{k}}{k}}+\ln{\left(1-x\right)}&=-\int_{0}^{x}{\frac{t^{n}}{1-t}\,\mathrm{d}t} \end{aligned}
Meaning, \begin{aligned}\left|\sum_{k=1}^{n}{\frac{x^{k}}{k}}+\ln{\left(1-x\right)}\right|=\left|\int_{0}^{x}{\frac{t^{n}}{1-t}\,\mathrm{d}t}\right|\leq\left|x\right|^{n}\int_{0}^{x}{\frac{\mathrm{d}t}{1-t}}\leq\left|x\right|^{n}\int_{0}^{1}{\frac{\mathrm{d}t}{1-t}}=\left|x\right|^{n}\ln{2}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}
And thus, $$ \sum_{n=1}^{+\infty}{\frac{x^{n}}{n}}=\lim_{n\to +\infty}{\sum_{k=1}^{n}{\frac{x^{k}}{k}}}=-\ln{\left(1-x\right)} $$
For $ x>\frac{1}{2} $, setting $ x\leftarrow\frac{x-1}{x}<1 $, we get : $$ \sum_{n=1}^{+\infty}{\frac{1}{n}\left(\frac{x-1}{x}\right)^{n}}=-\ln{\left(1-\frac{x-1}{x}\right)}=\ln{x} $$
