Endpoint of an interval of convergence

Substituting $x = a + R$ and $x = a - R$ turns the power series into two ordinary numerical series. The two results are independent, because the factor $(x-a)^n$ is $R^n$ at one end and $(-R)^n$ at the other, so one end keeps a fixed sign while the other alternates.

Two series with the same centre and the same radius can still have different intervals. The series $\sum_{n=1}^{\infty} \frac{x^n}{n}$ converges on $[-1, 1)$, because $x = -1$ gives the alternating harmonic series and $x = 1$ gives the harmonic series, while $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$ converges on the closed $[-1, 1]$, since $x = 1$ gives $\sum \frac{1}{n^2}$ and $x = -1$ gives that same series with alternating signs.

Term-by-term differentiation keeps the radius but can cost you an end. Differentiating $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$, which converges on $[-1, 1]$, gives $\sum_{n=1}^{\infty} \frac{x^{n-1}}{n}$, and that series diverges at $x = 1$.