# Radius of convergence for complex power series

I am supposed to find the radius of convergence for the complex power series $$\sum_{n=0}^{\infty}(-1)^n2^nz^{2n+2}$$ I know that the radius of convergence is calculated by $\frac{1}{R}=\limsup_{n\to\infty}|a_n|^{1/n}$, which in this case means $$\frac{1}{R}=\limsup_{n\to\infty}|(-2)^n|^{1/n}=2$$ $$\Rightarrow R=\frac{1}{2}$$ but the answer is supposed to be $R=\frac{1}{\sqrt{2}}$. Where did I go wrong? Is it because of the $z^{2n}$ in the series?

• That formula for $R$ is for $\sum a_n z^n$ not $\sum a_n z^{2n+2}$. – nullUser Feb 3 '16 at 1:00
• How do I calculate the radius for $\sum a_nz^{2n+2}$? Is there a formula, or do I have to manually find where it converges? – Matt G Feb 3 '16 at 1:05

Yes, you went wrong because of the $z^{2n}$. You're assuming that $a_n=(-1)^n2^n$. That's not so. In fact $a_{2n+2}=(-1)^n2^n$, while $a_k=0$ if $k$ is not of the form $k=2n+2$.
The best way to look at these things, in my opinion, is to forget that $|a_n|^{1/n}$ and know this: The radius of convergence is the supremum of the $r$ such that $|a_n|r^n$ is bounded.
So here you'd ask yourself when $2^nr^{2n+2}$ is bounded. Since $2^nr^{2n+2}=(2r^2)^nr^2$, it's bounded if and only if $2r^2\le1$. Which is if and only if $r<1/\sqrt 2$. So the radius of convergence is $1/\sqrt 2$.
$$\sum_{n=0}^\infty (-1)^n 2^n z^{2n+2} = z^2 - 2z^4 + 4z^6 - 8z^8 + 16z^{10} - \dotsb := \sum_{n=2}^\infty b_n z^n$$ where $b_n = 0$ if $n$ is odd and $(-2)^{n/2-1}$ if $n$ is even. Then the sequence $|b_n|^{1/n}$ alternates between $0$ and $2^{1/2-1/n}$ according to the parity of $n$, so the sequence of suprema converges to $\sqrt{2}$.
The calculation is simpler if one factors out $z^2$ initially, because then the series is equal to $$z^2\sum_{n=0}^\infty c_n z^n$$ where $c_n = (-2)^{n/2}$ for even $n$, and 0 otherwise.