Thm 37.1

For $\sum a_{k} x^{k} = f (x)$ we can prove

Thm 37.2

For uniform limit $lim f_{n}^{'} = g$ where $g$ is continue and $lim f_{n} = f$ point-wise, we can say $f^{'} = g$ .

For $lim f_{n} = f$ is uniform limit and $lim f_{n} = f$ in Metric Space $⟹$ $lim \int f_{n} (t) d t = \int f (t) d t$ is uniform limit.

R = [n lim sup n ∣ a_{n} ∣]^{- 1}

$(\forall r < R) f_{n} (x) = \sum_{k = 0}^{n} a_{k} x^{k}$ are uniformly Cauchy on $C [- r, r]$

$(\forall r < R) f_{n}$ uniformly converge to $f$ on $[- r, r]$ and $\sum a_{k} x^{k}$ convergent.

n lim sup n ∣ a_{n} ∣ = n lim sup n (k + 1) ∣ a_{n + 1} ∣ = n lim sup n \frac{∣ a _{n - 1} ∣}{k}

let $(a_{n})$ and $R = (lim_{n} sup n ∣ a_{n} ∣)^{- 1}$
Then on $(- R, R)$ , $f (x) = \sum_{k} a_{k} x^{k}$ is

\frac{1}{1 - x} = k = 0 \sum \infty x^{k}

Radius of convergent comes from complex inequality.

k lim sup k k = 1