In mathematics, the radius of convergence of a power series

[itex]f(z)=\sum_{n=0}^\infty c_n (z-a)^n,[itex]

where the center a and the coefficients cn are complex numbers (which may, in particular, be real numbers) is the nonnegative quantity r (which may be a real number or ∞) such that the series converges if

[itex]\left|z-a\right|

and diverges if

[itex]\left|z-a\right|>r.[itex]

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 radius of convergence is infinite if the series converges for all complex numbers z.

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## Existence and value of the radius of convergence

The radius of convergence can be found by applying the root test to the terms of the series. If

[itex]C = \limsup_{n\rightarrow\infty}\sqrt[n]{|c_n|}[itex]

then the radius of convergence is 1/C. If C = 0, then the radius of convergence is infinite, meaning that f is an entire function.

In many cases, the ratio test suffices. If the limit

[itex] L = \lim_{n\rightarrow\infty} \left| \frac{c_{n+1}}{c_n} \right| [itex]

exists, then the radius of convergence is 1/L. This limit is often easier to compute than the limit for C above. However, it may not exist, in which case one has to use the formula for C instead.

## Clarity and simplicity result from complexity

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 (non-removable) 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. See proof that holomorphic functions are analytic; the result stated above is a by-product of that proof.

### A simple warm-up example

The arctangent function of trigonometry can be expanded in a power series familiar to calculus students:

[itex]\arctan(z)=z-\frac{z^3}{3}+\frac{z^5}{5}-\frac{z^7}{7}+\cdots .[itex]

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:

[itex]\frac{d}{dz}\arctan(z)=\frac{1}{1+z^2}[itex]

and a zero appears in the denominator when z2 = − 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.

### A gaudier example

Consider this power series:

[itex]\frac{z}{e^z-1}=\sum_{n=0}^\infty B_n z^n[itex]

where the coefficients Bn 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 non-removable singularities are therefore located where the denominator is zero. We solve

[itex]e^z-1=0[itex]

by recalling that if z = x + iy and eiy = cos(y) + i sin(y) then

[itex]e^z = e^x e^{iy} = e^x(\cos(y)+i\sin(y)),[itex]

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 ez can be 1 only if ex 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 2πi or − 2πi. The distance from the center to either of those points is 2π. That is therefore the radius of convergence.de:Konvergenzradius

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