I have a series $S$ with general terms $a_n=\frac{(-1)^n(x-1)^n}{(2n-1)2^n}$, $n\ge 1$:
$$S = \sum_{n=1}^\infty \frac{(-1)^n(x-1)^n}{(2n-1)2^n}$$
Finding the ratio $\left|\frac{a_{n+1}}{a_n}\right|$ and then finding the limit of the ratio as $n\to\infty$, I find the limit to be $1$ and the interval to be $-1 \lt x \lt 3$. More declaratively, the interval is $\left|\frac{x−1}{2}\right| \lt 1$ which I've refined to what was said earlier.
I've read conflicting sites that state the radius $R$ of convergence is $\frac{1}{N}$, where $N$ is the limit as found earlier, but also that it's half the interval length.
Here's my work:
$$\begin{align} \left|\frac{a_{n+1}}{a_n}\right| &= \left|\frac{\frac{(-1)^{n+1}(x-1)^{n+1}}{(2(n+1)-1)2^{n+1}}}{\frac{(-1)^{n}(x-1)^{n}}{(2n-1)2^{n}}}\right| \\ &= \left|\frac{(-1)^{n+1}(x-1)^{n+1}(2n-1)(2^n)}{(-1)^n(x-1)^n(2(n+1)-1)(2^{n+1})}\right| \\ &= \left|\frac{(-1)(x-1)(2n-1)}{(2n+2-1)(2)}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times \left|\frac{2n-1}{2n+2}\right| \end{align}$$
Then, finding the limit $L$:
$$\begin{align} L &= \lim_{n\to\infty} \left(\left|\frac{-(x-1)}{2}\right| \times \left|\frac{2n-1}{2n+1}\right|\right) \\ &= \left|\frac{-(x-1)}{2}\right| \times \lim_{n\to\infty} \left|\frac{2n-1}{2n+1}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times \lim_{n\to\infty} \left|\frac{\frac{2n}{n}-\frac{1}{n}}{\frac{2n}{n}+\frac{1}{n}}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times \lim_{n\to\infty} \left|\frac{2-\frac{1}{n}}{2+\frac{1}{n}}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times \left|\frac{\lim_{n\to\infty} \left(2-\frac{1}{n}\right)}{\lim_{n\to\infty} \left(2+\frac{1}{n}\right)}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times \left|\frac{2}{2}\right| \\ &= \left|\frac{-(x-1)}{2}\right| \times 1 \\ &= \left|\frac{-(x-1)}{2}\right| \end{align}$$
Then I know my interval is $\left|\frac{-(x-1)}{2}\right| \lt 1$:
$$\left|\frac{-(x-1)}{2}\right| \lt 1 \\ -1 \lt \frac{x-1}{2} \lt 1 \\ -2 \lt x-1 \lt 2 \\ -1 \lt x \lt 3$$
If the limit found earlier is $1$, the radius would be $R = \frac{1}{1} = 1$, yet I've found the interval to be $(-1, 3)$, which would imply $R = 2$. Where have I made an error?