Radius and Interval of Convergence of Power Series BC only

Applying the ratio test to $\sum c_n(x-a)^{n}$ with $x$ fixed gives $L = \left|x-a\right|\lim\left|\frac{c_{n+1}}{c_n}\right|$, and solving $L < 1$ produces the radius directly, with $R = \infty$ when the coefficient ratio tends to zero, as for $e^{x}$, $\sin x$ and $\cos x$, and $R = 0$ when the series converges only at its centre. Substituting $x = a \pm R$ makes $L$ exactly $1$, because $\left|x-a\right| = R$ and the coefficient ratio tends to $\frac{1}{R}$, so the endpoints are precisely where the ratio test returns the inconclusive case 10.8 showed carries no information. Each endpoint must therefore be substituted and tested as its own numerical series, using a $p$-series or comparison argument, the alternating series test, or the nth term test, and the two endpoints are independent of each other.

All four combinations of endpoint behaviour occur, so no bracket can be assumed: $\sum x^{n}$ converges on $(-1, 1)$ with both endpoints failing the nth term test, $\sum \frac{x^{n}}{n}$ on $[-1, 1)$ since $x = -1$ gives the alternating harmonic series and $x = 1$ the harmonic series, and $\sum \frac{x^{n}}{n^{2}}$ on $[-1, 1]$ since both endpoints give a convergent $p$-series. Those three share a radius of $1$ and differ only by a power of $n$, which shows the radius and the endpoint behaviour to be independent. The radius is a single number and the interval is a set with two decided endpoints, so answering with one when the other was asked for is answering a different question; a complete interval answer contains the ratio test, the radius, a separate test at each endpoint, and the brackets those tests determine.