Region of Convergence of power series

Question

The power series $\sum_{n=0}^\infty 2^{-n} z^{2n} $ converges if
a)$|z|\le 2$
b)$|z|\lt 2$
c)$|z|\le\sqrt2$
d)$|z|\lt\sqrt 2$
I tried this problem,my answer is d).I am not sure whether it is correct or not.It can be c).My question is how to check convergence of the series on the boundary of region and it is depend on function.Can anyone explain why and how?

Answer 1

$$\rho =\limsup_{n\to\infty}\sqrt[n]{|a_n |}=\limsup_{n\to\infty}\sqrt[2n]{|2^{-n} |} =\frac{1}{\sqrt{2}}$$ Hence $R=\frac{1}{\rho}=\sqrt{2}.$ So the answer d) is correct.

Answer 2

Hint

When you face things such as

$$S=\sum_{n=0}^\infty a^{n} z^{bn}$$ just rewrite it as $$S=\sum_{n=0}^\infty (az^b)^{n}= \sum_{n=0}^\infty x^{n}$$ where $x=az^b$. So, you are always back to the same problem.

For your specific case, $a=\frac 12$ and $b=2$.

Edit

Let us make it more complex $$S=\sum_{n=0}^\infty a^{bn+c} z^{dn}=a^c\sum_{n=0}^\infty a^{bn} z^{dn}=a^c\sum_{n=0}^\infty \big(a^{b} z^{d}\big)^n$$ Now, define $x=a^b z^d$ to go back to the simple situation.