This is the second attempt from my previous question found at: Complex Function: Radius of convergence and convolutions.
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 are given by way of a proven lemma in the lecture notes $$\sum_{n=0}^{\infty}n a_n z^{n-1}$$ has the same radius of convergence as $\sum_{n=0}^{\infty}a_n z^n$
then for $$\sum_{n=1}n a_n z^{n-1} = 0 + \sum_{n=1}n a_n z^{n-1} = \sum_{n=0}n a_n z^{n-1} = f'(z)$$ and so we have it that $\sum_{n=1}n a_n z^{n-1}$ has radius of convergence R.
Onto the next one (This took me way longer than it should, simply because i couldn't remember whether i could factor out the 2 in the limsup)
we have for $h(z) = \sum_{n=0}^{\infty}h_{n}z^n = \sum_{n=0}^{\infty}a^{2}_{n}z^n$ then the radius of convergence is given by $$\frac{1}{\overline{R}} = \limsup_{n \longrightarrow \infty} |h_n|^{1/n}= \limsup_{n \longrightarrow \infty} |a^2_n|^{1/n} = \limsup_{n \longrightarrow \infty} |a_n|^{2/n}=\frac{1}{R^2}$$
How about this attempt? it's been way too long since i last worked with the limsup. thanks for taking the time.
sincerely