The radius of convergence of the geometric series is 1.
Let's explore why. A geometric series can be expressed as:
$$\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \dots$$
where a is the first term and r is the common ratio. This series converges if |r| < 1 and diverges if |r| ≥ 1.
To determine the radius of convergence, we look at the interval of convergence, which is the set of all r for which the series converges. In this case, the interval of convergence is -1 < r < 1, or in interval notation, (-1, 1).
The radius of convergence (R) is half the length of this interval. Since the length of the interval (-1, 1) is 2 (1 - (-1) = 2), the radius of convergence is 2/2 = 1.
In the context of a power series of the form:
$$\sum_{n=0}^{\infty} c_n(x - a)^n$$
the geometric series can be represented by setting c_n = 1 and a = 0, and letting x = r. Therefore the series becomes:
$$\sum_{n=0}^{\infty} x^n$$
The interval of convergence is then -1 < x < 1, centered at 0. The radius of convergence is still 1, representing the distance from the center (0) to either endpoint of the interval (-1 or 1).
