Radius of convergence of series with alternating coefficients

Question

I need to compute, with proof, the radius of convergence $R$ for the series $$\sum_{k=0}^\infty \left(2-(-1)^n\right)^n z^n,$$ which is similar to a geometric series, except that the terms alternate between $z^n$ and $(3z)^n$. Clearly, we can upper-bound the terms by $(3z)^n$, giving $R \ge \frac{1}{3}$. However, I'm not how to deal with the alternating behavior of the terms. Am I correct to assume that the $(3z)^n$ terms ultimately dominate and give $R = \frac{1}{3}$?

Answer 1

There are a couple ways to do this, including the formula suggested by Daniel Fischer. However, here is a more elementary way.

First of all to find the radius of convergence you can consider the absolute value of the terms rather than the complex value. So you look at

$$ \sum_{n=0}^\infty (2 + (-1)^n)^n|z|^n $$

Now that the terms are all positive, we can rearrange them or split them into two sums. Let's split the terms into two separate sums as follows: \begin{align*} &\quad \sum_{n=0}^\infty (2 + (-1)^{2n})^{2n}|z|^{2n} + \sum_{n=0}^\infty (2 + (-1)^{2n + 1})^{2n+1}|z|^{2n + 1} \\ &= \sum_{n=0}^\infty 3^{2n} |z|^{2n} + \sum_{n=0}^\infty |z|^{2n + 1} \\ \end{align*}

The first series converges when $\left||3z|^2\right| < 1$, i.e. when $|z| < \tfrac13$. And the second series converges when $|z| < 1$. If $|z|$ lies outside either of these two intervals, one of the two series diverges, which makes the sum diverge as well. Thus indeed the radius of convergence must be $1/3$.

The key point is to consider the absolute value of the series, because strictly positive series are much easier to manipulate. (If the two series were not both positive, for example, we would have trouble concluding that if one series diverged, the sum of the two must diverge as well.)