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I was solving for the radius of convergence and regions where there is uniform convergence for the following problems -

$1)$ $\sum_{n = 0}^{\infty} 4^{n(-1)^{n}} z^{2n}$ where $z$ is a complex number.

I was approaching this by root test but got stuck while calculating $\lim_{n \to \infty}|4^{(-1)^n}||z^{2}|$. How to proceed further?

Also after if I find out the radius of convergence how to proceed with the uniform convergence part?

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    Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see [meta](http://meta.math.stackexchange.com/questions/6464/posting-multiple-questions-as-one). – 2017-02-05
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    @MartinSleziak ok.. thanks , edited the post. – 2017-02-05

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To get the radius of convergence of the series $$\sum_{n = 0}^\infty 4^{n(-1)^n}z^{2n}$$ we consider the following sequence $(a_n)_{n \in \mathbb{N}}$ defined by $$a_n := 4^{n(-1)^n}z^{2n}$$ The root test yields $$\limsup_{n \to \infty} |a_n|^{1/n} = \limsup_{n \to \infty} 4^{(-1)^n}|z|^2 = 4|z|^2$$ Now we must have $$4|z|^2 < 1 \qquad \Leftrightarrow \qquad |z| < \frac{1}{2}$$ Hence we get the radius of convergence $\frac{1}{2}$. For the other part, we have the following result.

Proposition. Let $\rho$ be the radius of convergence of a power series $$\sum_{n = 1}^\infty a_n z^n$$ For any $0 < r < \rho$ the power series is uniformly convergent on $\{z \in \mathbb{C} : |z| \leq r\}$.

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    i think we should deal with the radius of convergence on which i was having doubt and then we can come to uniform convergence as for uniform convergence part we need the radius of convergence – 2017-02-05
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    As was already mentioned above, it is very unlikely that someone will answer four questions. – 2017-02-05
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    @TheGeekGeek , ok , just now did an edit. – 2017-02-05
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    @BAYMAX Fine. See my edited answer. Hope it helps. – 2017-02-05
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    ok as we have the radius of convergence as $\frac{1}{2}$ so can we take any real number which is less than $\frac{1}{2}$ and $>0$ as region for uniform convergence ? – 2017-02-05
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    Ok got it! thank you! – 2017-02-05
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    @BAYMAX Yes, exactly. – 2017-02-05