Power Series

** Defn.**
If {c

å_{n = 0}^{¥}
c_{n} (z-z_{0})^{n}.

(Here we use the convention that 0^{0} = 0.)

** Defn.**
For a sequence {c

**r:
= 1/R**

where **R : = lim sup _{n®¥}
a_{n}^{1/n}**.

** Theorem.** Consider
the formal power series å

- the series converges absolutely if |z-z
_{0}| < r and the limit function f is continuous on this open disk. - the series converges uniformly and absolutely to f on
each closed disk |z-z
_{0}| £ r, where 0 £ r < r, - the series diverges if |z-z
_{0}| > r.

*Proof.* We first
prove part (2). Part (1) will follow since each point in the disk |z-z

|| f_{n} ||_{¥}
£ M_{n}

where the sup norm is taken over the closed set |z-z_{0}|
£ r. But the series {M_{n}} satisfies
lim sup_{n®¥} M_{n}^{1/n}
= r/r. If we set T : = r/r,
then 0 £ T < 1,
so å_{n = 0}^{¥}
M_{n} 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: = lim

* Proof.* An application
of the ratio test applied for fixed z to the power series å

** Proposition.**
Suppose k is a fixed natural number, then the two series å

*Proof.* Just note
that

lim sup_{n®¥} (n^{k}
|c_{n}|)^{1/n}
= (lim sup_{n®¥} |c_{n}|^{1/n}
) (lim_{n®¥} n^{1/n})^{k}.
^{[¯]}

** Corollary.** The
series å

*Proof.* Let r
be the radius of convergence of å

** 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) = å_{n
= 0}^{¥} a_{n} (z-z_{0})^{n}

then a_{n} = f^{(n)}(z_{0})/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__:

- å
_{n = 0}^{¥}1/n (z-z_{0})^{n}and å_{n = 0}^{¥}1/n^{2}(z-z_{0})^{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. - å
_{n = 0}^{¥}n^{n}(z-z_{0})^{n}has radius of convergence equal to zero.

: lim sup__Details___{n®¥}a_{n}^{1/n}= lim_{n®¥}n = ¥. - The series
**E(z) : =**å_{n = 0}^{¥}1/n! z^{n}has radius of convergence equal to ¥. Note that on the real line E¢(x) = E(x). - The series
**S(z) : =**å_{n = 0}^{¥}(-1)^{n}/(2n+1)! z^{2n+1}has radius of convergence equal to ¥.

: As in the previous Corollary, apply the ratio test to the values f__Details___{n}(z) = (-1)^{n}/(2n+1)! z^{(2n+1)}**:**

lim_{n®¥} |f_{n+1}(z)/f_{n}(z)|
= lim_{n®¥} |(2n+3)(2n+2)|^{-1}|z^{2}|
= 0.

- The series
**C(z) : =**å_{n = 0}^{¥}(-1)^{n}/(2n)! z^{2n}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).