Let $(a_n), (b_n)$ be sequences in $\mathbb{C}$ such that for given $r>0$ we have
$$\lim \sup |a_n|^{1/n} \leq \frac1r \text{ and } \lim \sup |b_n|^{1/n} \leq \frac1r$$
I want to show that for $c\in \mathbb{C}$ $$\lim \sup |a_n + b_n|^{1/n} \leq \frac1r \text{ and } \lim\sup |c\;a_n|^{1/n} \leq \frac1r$$
Basically, am trying to prove that the set $\mathcal{D}_r = \{ \sum_{0}^{\infty} a_n z^n\} \; | \; \lim \sup |a_n|^{1/n} \leq \frac1r \} $ is a subspace of the vector space of formal power series.'
For $\lim\sup |c\;a_n|^{1/n} \leq \frac1r $ I used that $$\lim\sup |c\;a_n|^{1/n} = \lim \frac{|c\cdot a_{n+1}|}{|c\cdot a_n|} = \lim \frac{|a_{n+1}|}{|a_n|} = \lim \sup |a_n|^{1/n}\leq \frac1r \;\;\;\;\text{Is this correct?}$$
I have tried hard but I am unable to show $\lim \sup |a_n + b_n|^{1/n} \leq \frac1r $
Help would me much appreciated.
From the definitions $\limsup x_n = \underset{n \to \infty}\lim \underset{k \geqslant n}\sup x_k$ and $\liminf x_n = \underset{n \to \infty}\lim \underset{k \geqslant n}\inf x_k$, it is not difficult to prove directly that
$$\liminf x_n \limsup y_n \leqslant \limsup(x_ny_n) \leqslant \limsup x_n \limsup y_n $$
It then follows that if $(x_n)$ converges so that $\liminf x_n =\limsup x_n = \lim x_n$, we have
$$\limsup(x_ny_n) = \lim x_n \limsup y_n$$
If $|c| \neq 0$ we have $\lim |c|^{1/n} = 1$. Hence,
$$\limsup |c a_n|^{1/n} =\limsup \left(|c|^{1/n} |a_n|^{1/n}\right)= \lim |c|^{1/n} \limsup |a_n|^{1/n} = 1 \cdot \limsup |a_n|^{1/n} \leqslant \frac{1}{r}$$
If $c = 0$, then it is a trivial consequence that $0 =\limsup |c a_n|^{1/n} \leqslant \frac{1}{r}$.