Consider a power series \(\sum c_n (x-a)^n\). It could be the case that the series only converges at \(x=a\), and diverges at every other \(x\) value. It could also be the case, that the series converges for all \(x\) values. Let's assume that neither case is true.
Let \(z = x-a\), so now the series is \(\sum c_n z^n\). If it is not true that the series converges only when \(z = 0\), then there must an integer \(b\) such that the series converges when \(z = b\). If \(\sum c_n b^n\) converges, then \(\lim_{n \to \infty} c_n b^n = 0\) (contrapositive of the divergence test). By the definition of a limit, there must be a positive integer \(N\) such that \(|c_n b^n| \lt 1\) whenever \(n \ge N\). This means:
If \(|z| < |b|\), then \(\left| \frac{z}{b} \right| < 1\), so \(\sum \left| \frac{z}{b} \right| ^n\) is a convergent geometric series. Since \(|c_n z^n| < \left| \frac{z}{b} \right|^n\), then, by the comparison test, \(\sum |c_n z^n|\) is convergent, and thus \(\sum c_n z^n\) is also convergent. Thus we have proven, that if there exists some integer \(b \ne 0\) for which \(\sum c_n b^n\) converges, then \(\sum c_n z^n\) converges absolutely for all \(|z| < |b|\).
Now suppose that there is an integer \(d\) such that \(\sum c_n d^n\) diverges. If \(|z| > |d|\), then \(\sum c_n z^n\) diverges.
Let there be set \(S\) which includes all values of \(z\) where \(\sum c_n z^n\) would converge. If the series diverges whenever \(|z| > |d|\), then \(|d|\) must be an upper bound for the set, and \(-|d|\) must be a lower bound (i.e. all values of \(z\) in \(S\) must satisfy \(-d < z < d\)). This means that there must be some positive value \(R\) such that the series converges for all \(-R < z < R\). In other words, \(-R\) is the greatest upper bound for set \(S\) while \(R\) is the least upper bound. In such a case, \(R\) is called the radius of convergence.
Earlier we saw that if there exists some integer \(b \ne 0\) for which \(\sum c_n b^n\) converges, then \(\sum c_n z^n\) converges absolutely for all \(|z| < |b|\). If you choose any \(z\) in the set \(S\), you can also choose \(b\) such that \(z < b < R\), and show that the series converges absolutely for that \(z\). This fact that be used to show that the series \(\sum c_n z^n\) converges absolutely when \(|z| < R\).
One might think that the interval of convergence can be uneven. For example, instead of the series converging for values: \(-R < x-a < R\), the series can covergences for: \(-M < x-a < R\) where \(M \ne R\). According to the ratio test, a series \(\sum_{n=1}^\infty t_n\) is convergent if \(\lim_{n \to \infty} \left| \frac{t_{n+1}}{t_n} \right|<1\). Let's apply this test for the series \(\sum_{n=1}^\infty c_n (x-a)^n\):
Let the limit be \(\lim_{n \to \infty} \frac{c_{n+1}}{c_n} = L\):
Let \(L = R\):
This shows that if it's true that the series converges when the ratio between the terms as \(n\) goes to infinity is between -1 and 1, then it's also true that the series converges when \(x-a\) is between \(-R\) and \(R\) for some real number \(R\). Therefore, the interval is not uneven.