# Intuition behind radius of convergence for $\sum_{n=0}^\infty 2^n z^{2n}$

I've been trying to understand how to handle power series where they are not in the "standard" for of $c_n z^n$. Consider the example I put in the title: $\sum_{n=0}^\infty 2^nz^{2n}$. I was looking for some feedback / insight into understanding the radius of convergence for a complex power series.

For my example, we want:

$$\limsup_n |2^nz^{2n}|^{1/n} < 1$$

so that the series will converge. Re-arranging

$$\limsup_n|2^n|^{1/n}|z^2| = |z^2|\limsup_n|2|< 1$$

And we get the radius of convergence from $|z| < 1$, so in this case:

$$|z|^2 = \frac{1}{2} \rightarrow |z| < \frac{1}{\sqrt{2}} = R$$. However, this is just $\sqrt{R_o}$ where $R_o$ is the radius of convergence for $\sum_{n=0}^\infty 2^nz^{n}$. So any transformation to the $z^n$ gives radius of convergence of the new power series as some transformation to the radius of convergence of the old series? In my case, it makes sense that if we take away some of the terms of the sum of positive numbers $z^n$ then the power series still converges and within an even larger radius.

Am I on the right track here or completely misguided?

Yes, if the radius of convergence of a power series $f(z) = \sum_{n= 0}^{\infty} a_n z^n$ is $R$, then for any positive integer $k$ the radius of convergence of the power series $g(z) = \sum_{n=0}^{\infty} a_n z^{nk}$ is $R^{1/k}$. Since $g(z) = f(z^k)$, this follows directly from the definition of the radius of convergence: $g(z)$ converges if $|z^k| < R$ and diverges if $|z^k| > R$, hence converges if $|z| < R^{1/k}$ and diverges if $|z| > R^{1/k}$.

Incidentally, the root test is overkill here: $\sum_n 2^n z^{2n} = \sum_n (2z^2)^n$, so the series is geometric with geometric ratio $r = 2z^2$, hence it converges if and only if $|2z^2| = 2 |z|^2 < 1$.

In my case, it makes sense that if we take away some of the terms of the sum of positive numbers $z^n$ then the power series still converges and within an even larger radius. Am I on the right track here or completely misguided?

This doesn't sound so good to me. If all you were doing was removing some portion of the terms of the series, that would not necessarily change the radius of convergence -- it will iff passage to a subsequence makes $\limsup_n |a_n|^{1/n}$ smaller. In particular if $\lim_{n \rightarrow \infty} |a_n|^{1/n}$ exists then removing some of the terms (but keeping infinitely many, of course!) does not change the radius of convergence.

Instead the the exponents of $2^n$ and $z^{2n}$ do not match up...which is best explained by starting with $\sum_n 2^n z^n$ and making the substitution $z \mapsto z^2$, as I did above. You should think about the similar example $\sum_n 2^{-n} z^n$. This time the radius of convergence is $2$, but after $z \mapsto z^2$, the radius of convergence of $\sum_n 2^{-n} z^{2n}$ is $\sqrt{2}$: that's smaller, not bigger.

Finally, none of this is any different for complex power series than for real ones. This could be a helpful remark if you have prior familiarity with real power series and are just starting to think about complex power series (otherwise not: it's not any different!).

• Thanks for the clarification! This helped a lot – student_t Sep 27 '17 at 22:35