Let $R>0$ be the radius of convergence of $\sum_{n=0}^{\infty}a_nz^{n}$, then what is the radii of convergence of $\sum_{n=0}^{\infty} a^{2}_{n}z^{n}$ and $\sum_{n=1}^{\infty}n a_nz^{n-1}$?
We'll do the easy one first.
we know that for an analytic function $f(z)=\sum_{n=0}^{\infty} a_n z^n$ with >radius of convergence R then $\exists$ a function $f'(z) = \sum_{n=0}^{\infty}n >a_n z^{n-1}$
Comparing the two functions $g(z) = \sum_{n=1}^{\infty}n a_n z^{n-1}$ and $f'(z)$ >we have
$$\lim_{n \longrightarrow \infty}|f'(z)-g(z)| = \lim_{n \longrightarrow \infty} >|\sum_{n=0}^{\infty}n a_n z^{n-1} – \sum_{n=1}^{\infty}n a_n z^{n-1}|$$
$$\lim_{n \longrightarrow \infty} |0+\sum_{n=1}^{\infty}n a_n z^{n-1} – >\sum_{n=1}^{\infty}n a_n z^{n-1}| = 0~\forall n\in \mathbb{N}$$
$$\implies$$
$$g(z) = f'(z)$$ and so $g(z)$ has Radius of convergence R, the same as $f(z)$now the slightly more interesting one.
define $h(z) = \sum_{n=0}^{\infty} a^{2}_{n}z^{n}$ which we can clearly define to >be a cauchy product and so $h(z)$ is just the convolution of $f(z)^2$
since via abel's lemma:
$$\text{if }~\{a_n\} \subset \mathbb{C} \text{ and }\sum a_n \text{ converge, then >} \lim_{x \longrightarrow 1^{-}}\sum_{n=0}^{\infty} a_nx^{n} = \sum_{n=0}^{\infty} >a_n$$and the convolution lemma:
If the complex series $\sum a_n$ and $\sum b_n$ converge absolutely, then so does >their convolution $c_n = \sum_{i=0}^{n}a_ib_{n-i}$ and $\sum_{n=0}^{\infty}c_n = >\sum_{n=0}^{\infty}a_n \sum_{n=0}^{\infty}b_n$
applying these two lemmas we have
$$\sum_{n=0}^{\infty}a^2_{n} = \lim _{n \longrightarrow 1^{-}} \sum_{n=0}^{\infty}a^{2}_{n}z^{n} = \lim _{n \longrightarrow 1^{-}} \left(\sum_{n=0}^{\infty}a_{n}z^{n}\right)\left(\sum_{n=0}^{\infty}a_{n}z^{n}\right) = \left(\sum_{n=0}^{\infty}a_{n}\right)\left(\sum_{n=0}^{\infty}a_{n}\right) = f(z)f(z)$$
and so we can say that the radius of convergence of $h(z)$ is $\overline{R} \geq \min\{R,R\} \implies \overline{R} \geq R$we know that $$\frac{1}{R} = \limsup_{n \longrightarrow \infty} |a_{n}|^{1/n}$$
letting $h_n$ be the co-coefficients of $h(z) = \sum_{n=0}^{\infty} h_{n}z^{n} = \sum_{n=0}^{\infty} a^{2}_{n}z^{n}$then $$\limsup_{n \longrightarrow \infty}|h_{n}|^{1/n} = \limsup_{n \longrightarrow \infty}|a_{n}^{2}|^{1/n}= \limsup_{n \longrightarrow \infty}>(|a_{n}||a_n|)^{1/n}$$
for $a_n = u(x,y)+iv(x,y), |a_n| \in \mathbb{R}$
$$\limsup_{n \longrightarrow \infty}(|a_{n}||a_n|)^{1/n} = \limsup_{n \longrightarrow \infty}(|a_{n}|^{1/n}|a_n|^{1/n}) = (\limsup_{n \longrightarrow \infty}|a_{n}|^{1/n})(\limsup_{n \longrightarrow \infty}|a_{n}|^{1/n}) = \frac{1}{R^2}$$ Which is fine given that it is not of the form $0 \cdot \infty$ and we note that
$$\overline{R} = R^{2} \geq R$$so for $h(z)$ we find by squaring the co-coefficients of the power series we have >also squared the radius of convergence.<\strike>
Have i gone wrong anywhere?
Thanks for taking the time to read this.
Sincerely
The above answer is Wrong, Please see my second attempt at Complex Function: Radius of convergence Pt 2: Complex Boogaloo
Best Answer
You have gone wrong, yes.
In the case of the first series, what do you mean by $\displaystyle\lim_{n\to\infty}\bigl\lvert f'(z)-g(z)\bigr\rvert$? There is no $n$ in $\bigl\lvert f'(z)-g(z)\bigr\rvert$ and therefore, for each $z\in\mathbb C$, $\displaystyle\lim_{n\to\infty}\bigl\lvert f'(z)-g(z)\bigr\rvert=\bigl\lvert f'(z)-g(z)\bigr\rvert$.
You can prove that both series have the same redius of convergence by proving that$$\limsup_{n\to\infty}\sqrt[n]{\lvert a_n\rvert}=\limsup_{n\to\infty}\sqrt[n]{n\lvert a_{n-1}\rvert}.$$
Concerining your other problem, it is not true that$$f(z)=\sum_{n=0}^\infty a_nz^n\implies f^2(z)=\sum_{n=0}^\infty{a_n}^2z^n.$$For instance,$$\frac1{1-z}=\sum_{n=0}^\infty z^n=\sum_{n=0}^\infty1\times z^n\text{ but }\frac1{1-z^2}\neq\sum_{n=0}^\infty1^2z^n=\sum_{n=0}^\infty z^n.$$In order to solve this problem, you can use again the $\limsup$ approach.