# Monday January 13th

Recall that $\sum z_k$ converges iff $s_n = \sum_{k=1}^n z_k$ converges.

Lemma
: Absolute convergence implies convergence.

The most interesting series: $f(z) = \sum a_k z^k$, i.e. power series.

Lemma (Divergence)
: If $\sum z_k$ converges, then $\lim z_k = 0$.

Corollary
: If $\sum z_k$ converges, $\theset{z_k}$ is uniformly bounded by a constant $C > 0$, i.e. $\abs{z_k} < C$ for all $k$.

**Proposition:**
If $\sum a_k z_k$ converges at some point $z_0$, then it converges for all $\abs z < \abs z_0$.

Note that this inequality is necessarily strict.
For example, $\sum \frac{z^{n-1}}{n}$ converges at $z=-1$ (alternating harmonic series) but not at $z=1$ (harmonic series).

Proof
: 	Suppose $\sum a_k z_1^k$ converges.
	The terms are uniformly bounded, so $\abs{a_k z_1^k} \leq C$ for all $k$.
	Then we have $$\abs {a_k} \leq C/\abs{z_1}^k$$, so if $\abs z < \abs{z_1}$ we have $$\abs{a_k z^k} \leq \abs{z}^k \frac{C}{\abs{z_1}^k} = C (\abs{z} / \abs{z_1} )^k.$$
	So if $\abs{z} < \abs{z_1}$, the parenthesized quantity is less than 1, and the original series is bounded by a geometric series.
	Letting $r = \abs{z} / \abs{z_1}$, we have

	\begin{align*}
	\sum \abs{a_k z^k} \leq \sum c r^k = \frac{c}{1-r}
	,\end{align*}

	and so we have absolute convergence.

Exercise (future problem set)
: Show that $\sum \frac 1 k z^{k-1}$ converges for all $\abs{z} = 1$ except for $z = 1$.
	(Use summation by parts.)

Definition (Radius of Convergence)
: The *radius of convergence* of a series is the real number $R$ such that $f(z) = \sum a_k z^k$ converges precisely for $\abs z < R$ and diverges for $\abs z > R$.

	We denote a disc of radius $R$ centered at zero by $D_R$.
	If $R=\infty$, then $f$ is said to be *entire*.

Proposition
: Suppose that $\sum a_k z^k$ converges for all $\abs{z} < R$.
	Then $f(z) = \sum a_k z^k$ is continuous on $D_R$, i.e. using the sequential definition of continuity, $\lim_{z\to z_0} f(z) = f(z_0)$ for all $z_0 \in D_R$.

Recall that $S_n(z) \to S(z)$ uniformly on $\Omega$ iff $\forall \varepsilon > 0$, there exists a $M\in \NN$ such that $$n> M \implies \abs{S_n(z) - S(z)} < \varepsilon$$ for all $z\in \Omega$

Note that arbitrary limits of continuous functions may not be continuous.
Counterexample: $f_n(x) = x^n$ on $[0, 1]$; then $f_n \to \delta(1)$.
This uniformly converges on $[0, 1-\varepsilon]$ for any $\varepsilon > 0$.

Exercise
: Show that the uniform limit of continuous functions is continuous.
	
	> Hint: Use the triangle inequality.

Proof (of proposition)
: Write $f(z) = \sum_{k=0}^N a_k z^k + \sum_{N+1}^\infty a_k z^k \definedas S_N(z) + R_N(z)$.
	Note that if $\abs{z} < R$, then there exists a $T$ such that $\abs{z} < T < R$ where $f(z)$ converges uniformly on $D_T$.

	> Check!

	We need to show that $\abs{R_N(z)}$ is uniformly small for $\abs{z} < s < T$.
	Note that $\sum a_k z^k$ converges on $D_T$, so we can find a $C$ such that $\abs{a_k z^k} \leq C$ for all $k$.
	Then $\abs{a_k} \leq C/T^k$ for all $k$, and so

	\begin{align*}
	\abs{\sum_{k=N+1}^\infty a_k z^k}
	&\leq \sum_{k=N+1}^\infty \abs{a_k} \abs{z}^k \\
	&\leq \sum_{k=N+1}^\infty  (c/T^k) s^k \\
	&= c\sum \abs{s/T}^k \\
	&= c \frac{r^{N+!}}{1-r}
	&= C \varepsilon_n \to 0
	,\end{align*}

	which follows because $0 < r = s/T < 1$.

	So $S_N(z) \to f(z)$ uniformly on $\abs{z} < s$ and $S_N(z)$ are all continuous, so $f(z)$ is continuous.

There are two ways to compute the radius of convergence:

- Root test: $\lim_k \abs{a_k}^{1/k} = L \implies R = \frac 1 L$.
- 
- Ratio test: $\lim_k \abs{a_{k+1} / a_k} = L \implies R = \frac 1 L$.

As long as these series converge, we can compute derivatives and integrals term-by-term, and they have the same radius of convergence.