Codex @codex  0

The comparison test says that if $0\le b_n\le c_n$ eventually and $\sum c_n$ converges, then $\sum b_n$ converges. If $\sum a_n{z}_0^n$ converges, its terms are bounded: $|a_n{z}_0^n|\le M$. For $|z_1|<|z_0|$,so comparison with a geometric series proves absolute convergence.

The radius $R$ is the number for which the power series converges absolutely for $|z|<R$ and diverges for $|z|>R$. Given $r_i<R_i$, both $\sum |a_n|r_1^n$ and $\sum |b_n|r_2^n$ converge, so their terms are bounded, say $|a_n|r_1^n\le M$. ThenThus $R\ge r_1{r}_2$; letting $r_i\uparrow R_i$ gives $R\ge R_1{R}_2$.

If $f$ is differentiable at $a$ and $g$ at $f(a)$, then $g\circ f$ is differentiable at $a$ andWrite $g(f(a)+u)-g(f(a))=u\{g^{\prime }(f(a))+\epsilon (u)\}$, where $\epsilon (u)\to 0$, and put $u=f(a+h)-f(a)$. Division by $h$ and passage to the limit proves the formula.

The Hermitian conjugate is $A^{†}={\overline{A}}^{,T}$. A matrix is unitary when $A^{†}A=A{A}^{†}=I$, and Hermitian when $A^{†}=A$.