Is the radius of convergence related to the ratio limit or half of the interval of convergence? 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?

Note that a power series takes the form
$$\sum_{n=0}^\infty a_n(x-x_0)^n$$
In your case you have
$$a_n=\begin{cases}\frac{(-1)^n}{(2n-1)2^n}&n\ne0\\0&n=0\end{cases}\qquad x_0=1$$
If you calculate the limit you call $N$ we get
$$N=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac12$$
So the radius of convergence is $R=1/N=2$ and hence the interval of convergence is
$$x\in(x_0-R,x_0+R)=(-1,3)$$
as expected.
