Radius of convergence
In mathematics, the radius of convergence of a power series
where the center
a and the coefficients
c_{n} are
complex numbers (which may, in particular, be
real numbers) is the nonnegative quantity
r such that the series converges if
and diverges if
In other words, the series converges if
z is close enough to the center. The radius of convergence specifies how close is close enough.
The word "nonnegative quantity" is used where "nonnegative number" would not quite have sufficed: for some power series, the radius of convergence is ∞.
The existence of that quantity follows from the ratio test for convergence of series of complex numbers. According to the ratio test,
the series
converges if
and diverges if
In particular, if the limit of the sequence

t_{n+1}/
t_{n}  exists, then the series converges if that limit is less than 1 and diverges if it is more than 1. (Both convergent and divergent series exist for which the limit of that sequence is exactly 1; hence the test is inconclusive in that case.) Applied to the power series, the limiting ratio simplifies to
This proves existence provided that this last limit exists. If it does not, other tests, such as the root test, may serve. But proof of existence that need not rely on this proviso follows from a theorem of complex analysis stated in the next section.
One of the best examples of clarity and simplicity following from thinking about complex numbers where confusion would result from thinking about real numbers is this theorem of complex analysis:
 The radius of convergence is always equal to the distance from the center to the nearest point where the function f has a (nonremovable) singularity; if no such point exists then the radius of convergence is infinite.
The nearest point means the nearest point in the complex plane, not necessarily on the real line, even if the center and all coefficients are real.
The arctangent function of trigonometry can be expanded in a power series familiar to calculus students:
It is easy to apply the ratio test in this case to find that the radius of convergence is 1. But we can also view the matter thus:
and a zero appears in the denominator when
z^{2} = − 1, i.e., when
z =
i or −
i. The center in this power series is at 0. The distance from 0 to either of these two singularities is 1. That is therefore the radius of convergence.
Consider this power series:
where the coefficients B_{n} are the Bernoulli numbers. It may be cumbersome to try to apply the ratio test to find the radius of convergence of this series. But the theorem of complex analysis stated above quickly solves the problem. At
z = 0, there is in effect no singularity since
the singularity is removable. The only nonremovable singularities are therefore located where the denominator is zero. We solve
by recalling that if
z =
x +
iy then
and then take
x and
y to be real. Since
y is real, the absolute value of cos(
y) +
i sin(
y) is necessarily 1. Therefore, the absolute value of
e^{z} can be 1 only if
e^{x} is 1; since
x is real, that happens only if
x = 0. Therefore we need
cos(
y) +
i sin(
y) = 1. Since
y is real, that happens only if cos(
y) = 1 and sin(
y) = 0, so that
y is an integral multiple of 2π. Since the real part
x is 0 and the imaginary part
y is a nonzero integral multiple of 2π, the solution of our equation is
 z = a nonzero integral multiple of 2πi.
The singularity nearest the center (the center is 0 in this case) is at 2π or − 2π. The distance from the center to either of those points is 2π. That is therefore the radius of convergence.