$\arctan(z)=\sum\limits_{n=0}^\infty \frac{(-1)^n}{2n+1} z^{2n+1}$ from $(\arctan(z))'=\frac{1}{1+z^2}=\sum\limits_{n=0}^\infty (-1)^n z^{2n}$

Question

I'm at Kreyszig - "Advanced Engineering Mathematics" 10th ed. - sec. "15. Power Series, Taylor Series" - example 6. It finds the Maclaurin series of $f(z)=\arctan(z)$ by integrating

\begin{align*} f'(z)=\frac{1}{1+z^2}=\sum\limits_{n=0}^\infty (-1)^n z^{2n} && |z|<1 \tag{1} \end{align*}

term by term and, quote, 'using $f(0)=0$'.

Question: Why we can say that $\sum\limits_{n=0}^\infty \frac{(-1)^n}{2n+1} z^{2n+1}$ converges to the integral of $\frac{1}{1+z^2}$ (in $|z|<1$), given that it never mentioned that I can "switch the integral sign and the summation sign" to get the sum?

Background: up to now it proved that (I try to be concise so excuse me for highly informal math below):

1. Continuity: if $f=\sum a_n (z-z_0)^n$, then $f$ is continuos at $z_0$
2. Uniqueness: if $f$ has a power series, it's unique
3. Termwise differentiation: the series of derived terms has the same radius of convergence of the original series.
4. Termwise integration: the series of integrated terms has the same radius of convergence of the original series.
5. if $f$ has a power series, then $f$ is analytic and the series of derived terms is equal to $f'$
6. if $f$ is analytic, then it has a power series (the Taylor's series).

while chap. 14 was on Complex integration

Thanks, Luca

Answer 1

One of the main results about power series is that they converge UNIFORMLY on every compact subset of the disk of convergence (if you do not know what a compact subset is, just substitute "on any concentric disk of smaller radius").

Any uniformly convergent series can be integrated term-by term.
This is from calculus, but it is very easy to prove, once you understand the definition of uniform convergence. The only thing about integral used here is that $|\int_a^b f(z)dz|\leq \epsilon(b-a)$ whenever $|f(z)|\leq \epsilon$ for $z\in[a,b]$.

Answer 2

During the weekend I figured it out. Being $f(z)=\arctan(z)$, we are given

\begin{align*} & f'(z)=\frac{1}{1+z^2} \\ & f(0)=0 \end{align*}

and we already proved that $f(z)$ is analytic in $|z|<1$. For theorem 6. (see the question) it has a Taylor's series and the coefficients are

\begin{equation*} a_n=\frac{f^n(0)}{n!} \end{equation*}

so I can use the Taylor's series of $\frac{1}{1+z^2}$ (see $(1)$ in the question) to compute the coefficients $a_n$.

\begin{eqnarray*} f^1(z)=\sum\limits_{n=0}^\infty (-1)^n z^{2n} \\ f^2(z)=\sum\limits_{n=1}^\infty (-1)^n 2n z^{2n-1} \\ f^3(z)=\sum\limits_{n=1}^\infty (-1)^n 2n(2n-1) z^{2n-2} \\ f^4(z)=\sum\limits_{n=2}^\infty (-1)^n 2n...(2n-2) z^{2n-3} \\ f^5(z)=\sum\limits_{n=2}^\infty (-1)^n 2n...(2n-3) z^{2n-4} \\ f^6(z)=\sum\limits_{n=3}^\infty (-1)^n 2n...(2n-4) z^{2n-5} \\ f^7(z)=\sum\limits_{n=3}^\infty (-1)^n 2n...(2n-5) z^{2n-6} \\ f^8(z)=\sum\limits_{n=4}^\infty (-1)^n 2n...(2n-6) z^{2n-7} \\ f^9(z)=\sum\limits_{n=4}^\infty (-1)^n 2n...(2n-7) z^{2n-8} \\ \vdots \\ % guess \left\lbrace \begin{array}{ll} f^{2l}(z)=\sum\limits_{n=l}^\infty (-1)^n \frac{(2n)!}{(2n-2l+1)!} z^{2n-2l+1} & \text{even derivative}\\ \hline \begin{array}{l} f^{2l+1}(z)=\sum\limits_{n=l}^\infty (-1)^n \frac{(2n)!}{(2n-(2l+1)+1)!} z^{2n-(2l+1)+1}=\\ =\sum\limits_{n=l}^\infty (-1)^n \frac{(2n)!}{(2n-2l)!} z^{2n-2l} \end{array} & \text{odd derivative} \end{array} \right\rbrace & \text{guess}\\ % induction \left\lbrace \begin{array}{ll} \begin{array}{l} f^{2l+1}(z)=\sum\limits_{n=l}^\infty (-1)^n \frac{(2n)!(2n-2l+1)}{(2n-2l+1)!} z^{2n-2l}=\\ =\sum\limits_{n=l}^\infty (-1)^n \frac{(2n)!}{(2n-2l)!} z^{2n-2l} \end{array} & \text{got odd - ok}\\ \hline \begin{array}{l} f^{2l+2}(z)=\sum\limits_{n=l+1}^\infty (-1)^n \frac{(2n)!(2n-2l)}{(2n-2l)!} z^{2n-2l-1}=\\ =\sum\limits_{n=l+1}^\infty (-1)^n \frac{(2n)!}{(2n-2l-1)!} z^{2n-2l-1}=\\ =\sum\limits_{n=l+1}^\infty (-1)^n \frac{(2n)!}{(2n-2l-2+1)!} z^{2n-2l-2+1}=\\ =\sum\limits_{n=l+1}^\infty (-1)^n \frac{(2n)!}{(2n-2(l+1)+1)!} z^{2n-2(l+1)-1} \end{array} & \text{got next even - ok} \end{array} \right\rbrace & \text{induction} \end{eqnarray*}

Now we can compute $a_n=f^n(0)/n!$,

\begin{eqnarray*} & a_{2l}=\frac{f^{2l}(0)}{2l!}=\frac{1}{(2l)!}\sum\limits_{n=l}^\infty (-1)^n \frac{(2n)!}{(2n-2l+1)!} 0^{2n-2l+1}=0 \\ & a_{2l+1}=\frac{f^{2l+1}(0)}{(2l+1)!}=\sum\limits_{n=l}^\infty (-1)^n \frac{(2n)!}{(2n-2l)!} 0^{2n-2l}=(-1)^l \frac{(2l)!}{(2l+1)!(2l-2l)!}=\frac{(-1)^l}{2l+1} \end{eqnarray*}

so, given that $f(0)=0$ and so $a_0=0$ we have built all the coefficients of the Taylor's series of $f(z)$,

\begin{eqnarray*} f(z)=\arctan(z)=a_0+\sum\limits_{l=0}^\infty \frac{(-1)^l}{2l+1} z^{2l+1} \end{eqnarray*}

For theorem 2. (uniqueness, see question) I can say that this the unique power series that represents $\arctan(z)$.