The ratio test finds the radius and cannot see the endpoints BC only

A power series $\sum c_n (x-a)^{n}$ converges for $x$ within some distance $R$ of $a$ and diverges beyond it. To find $R$, apply the ratio test treating $x$ as fixed:

$$L = \lim_{n \to \infty}\left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^{n}}\right| = \left|x-a\right| \lim_{n \to \infty}\left|\frac{c_{n+1}}{c_n}\right|.$$

Convergence needs $L < 1$, which is an inequality in $\left|x-a\right|$, and solving it gives the radius directly. Three outcomes:

1. $0 < R < \infty$: the usual case. Converges on an interval, endpoints still unknown.
2. $R = \infty$: converges for every $x$. This happens when the coefficient ratio tends to $0$, as for $e^{x}$, $\sin x$ and $\cos x$, and there are no endpoints to test.
3. $R = 0$: converges only at $x = a$. The series is useless as a function but the answer is still an interval, the single point $\{a\}$.

Note that the ratio test is being used here as a tool with a variable in it, which is what makes it produce a region rather than a verdict. That is the whole reason it is the right tool for power series.

Substitute $x = a + R$ into the limit above. By construction $\left|x-a\right| = R$ and $\lim\left|\frac{c_{n+1}}{c_n}\right| = \frac{1}{R}$, so

$$L = R \cdot \frac{1}{R} = 1.$$

The same happens at $x = a - R$. The endpoints are exactly the values where the ratio test returns its inconclusive case, and 10.8 proved that case carries no information whatever. This is not a limitation to be worked around with a sharper computation; it is a fact about the boundary.

So each endpoint has to be handled as its own numerical series. Substitute the value, see what series you get, and test it with whatever fits:

1. A $p$-series or something like one: 10.5, or limit comparison from 10.6.
2. Alternating: 10.7's test.
3. Terms not tending to zero: 10.3 settles it immediately, and this is common at endpoints.

The two endpoints are independent, and testing one tells you nothing about the other. They frequently disagree, which is the entire reason four different interval shapes exist.

Two endpoints, each converging or not, give four possibilities, and every one of them is realised by a standard series:

1. Neither converges: $(a-R,\, a+R)$. Example: $\sum x^{n}$, with $R = 1$ and both endpoints giving $\sum (\pm 1)^{n}$, whose terms do not vanish.
2. Left only: $[a-R,\, a+R)$. Example: $\sum \frac{x^{n}}{n}$, where $x = -1$ gives the convergent alternating harmonic series and $x = 1$ gives the divergent harmonic series.
3. Right only: $(a-R,\, a+R]$. The mirror image, produced by a series whose signs are arranged the other way.
4. Both converge: $[a-R,\, a+R]$. Example: $\sum \frac{x^{n}}{n^{2}}$, where both endpoints give a convergent $p$-series with $p = 2$.

Since all four occur, no bracket can be assumed. Writing an interval with guessed endpoints is not a small omission: it is a claim about two separate convergence questions that were never asked.

The three examples above share a radius of $1$ and differ only in a power of $n$ in the denominator, which is a compact demonstration that the radius and the endpoint behaviour are genuinely independent pieces of information.

The other half of the code is answering the wrong question. They are different objects:

1. The radius $R$ is a single non-negative number, a distance.
2. The interval of convergence is a set, with two endpoints and a decision about each.

"$R = 2$" is not an interval, and "$(1, 5)$" is not a radius. When a question asks for the interval, the radius is an intermediate result and the endpoints are the remaining work; when it asks for the radius, testing the endpoints is effort spent on something not requested.

A complete interval answer therefore has four parts: the ratio test, the radius, a separate test at each endpoint, and the interval written with the correct brackets. Omitting the third is what CA18 records, and the resulting answer looks finished, which is why it survives to be marked.