ANALYSIS II
Power Series

Defn.  If {cn} is a sequence of real or complex numbers, and z0 is a fixed scalar, then define the formal power series for the sequence about the point z0 by

ån = 0¥ cn (z-z0)n.

(Here we use the convention that 00 = 0.)

Defn.  For a sequence {cn} of real or complex numbers, define the radius of convergence of the formal power series by

r: = 1/R

where R : = lim supn®¥  an1/n.

Theorem. Consider the formal power series ån = 0¥ cn (z-z0)n, then

1. the series converges absolutely if |z-z0| < r and the limit function f is continuous on this open disk.
2. the series converges uniformly and absolutely to f on each closed disk |z-z0| £ r, where 0 £ r < r,
3. the series diverges if |z-z0| > r.
4. Proof. We first prove part (2). Part (1) will follow since each point in the disk |z-z0| < r (or interval -r < z-z0 < r, if we are considering the reals), lies in a closed disk |z-z0| £ r for some 0 < r < r. To prove (2), we suppose that 0 < r < r and use the Weierstrass M-test. Set fn(z) : = cn (z-z0)n and Mn : = |cn| rn, then

|| fn ||¥ £ Mn

where the sup norm is taken over the closed set |z-z0| £ r. But the series {Mn} satisfies lim supn®¥  Mn1/n = r/r. If we set T : = r/r, then 0 £ T < 1, so ån = 0¥ Mn converges by the n-th root test. Similar reasoning, but using the divergence criteria from the n-th root test verifies part (3) of the theorem.   [¯]

Corollary.  If b: = limn®¥ |cn/cn+1| exists (in the extended sense), then this limit is equal to the radius of convergence of the power series ån = 0¥ cn (z-z0)n.

Proof. An application of the ratio test applied for fixed z to the power series ån = 0¥ cn (z-z0)n shows that limn®¥ |cn+1 (z-z0)n+1/cn (z-z0)n| = limn®¥ |cn+1|/|cn| |z-z0| = |z-z0|/b. Hence the series converges if |z-z0| < b and diverges for |z-z0| > b.   [¯]

Proposition.  Suppose k is a fixed natural number, then the two series ån = 0¥ cn (z-z0)n and ån = 0¥ nk cn (z-z0)n have the same radius of convergence.

Proof. Just note that

lim supn®¥  (nk |cn|)1/n = (lim supn®¥  |cn|1/n ) (limn®¥  n1/n)k.   [¯]

Corollary. The series ån = 1¥ n cn (z-z0)n-1 and ån = 0¥ cn (z-z0)n have the same radius of convergence.

Proof. Let r be the radius of convergence of ån = 0¥ cn (z-z0)n. We show that the second power series converges for |z-z0| < r and diverges for |z-z0| > r. Hence r is also the radius of convergence of the second function. To determine convergence of ån = 1¥ n cn (z-z0)n-1 we consider the expression R1 = lim supn®¥  (n |cn| |z-z0|n-1)1/n. But we may rewrite this as R1 = lim supn®¥  (n |cn|)1/n |z-z0|1-1/n. But limn®¥ |z-z0|(1-1/n) = 1 holds, by recalling that for a > 0 the function ax = exp(a log(x)) is continuous at x = 1.   [¯]

Theorem.  Power series are infinitely differentiable functions inside their disk of convergence. To obtain the derivative of a power series, one may differentiate it term by term. On each compact subset of that disk, the derived series converges uniformly and absolutely to the derivative of the function to which the power series converges. Moreover, if

f(z) = ån = 0¥ an  (z-z0)n

then an = f(n)(z0)/n! for all n.

Proof. Combine the results of this section with the interchange of limits results for sequences of functions (applied to the sequence of partial sums).   [¯]

Examples:

1. ån = 0¥ 1/n (z-z0)n and ån = 0¥ 1/n2 (z-z0)n both have radius of convergence equal to 1. This shows that in general the series may converge or diverge on its circle of convergence.
2. ån = 0¥ nn (z-z0)n has radius of convergence equal to zero.
Details:  lim supn®¥ an1/n = limn®¥ n = ¥.
3. The series E(z) : = ån = 0¥ 1/n!  zn has radius of convergence equal to ¥. Note that on the real line E¢(x) = E(x).
4. The series S(z) : = ån = 0¥ (-1)n/(2n+1)!  z2n+1 has radius of convergence equal to ¥.
Details:  As in the previous Corollary, apply the ratio test to the values fn(z) = (-1)n/(2n+1)!  z(2n+1):

limn®¥ |fn+1(z)/fn(z)|  =  limn®¥ |(2n+3)(2n+2)|-1|z2 =  0.

1. The series C(z) : = ån = 0¥ (-1)n/(2n)!  z2n also has radius of convergence equal to ¥. Note that on the real line S¢(x) = C(x) and C¢(x) = -S(x). Moreover, E(ix) = C(x) + i S(x).

Robert Sharpley March 31 1998