Convergence for power series $\sum_{n=0}^{\infty}z^n(5+e^{i\pi n})^n$?

Question

Consider the power series

$$\sum_{n=0}^{\infty}z^n(5+e^{i\pi n})^n \tag{1}.$$

Identifiying $e^{i\pi n}$ as $(-1)^n$ for all integers $n$, highlights the problem with using the ratio test or the root test for convergence of series, since $(-1)^n$ does not converge, right?

How can the convergence be determined for this power series? Does it exist?

Does it exist for other than $z=0$? Can it be determined by setting $(-1)^n=-1$ for all $n$ and therefore calculate the "smallest" radius or something like that?

Answer 1

Let's set $a_n=(5+(-1)^n)z^n$.

We have $|4z|^n\le|a_n|\le |6z|^n$ so the series converges for sure whenever $|z|<\frac 16$ and diverges for sure whenever $|z|>\frac 14$

What about values in the ring $\frac 16\le |z|\le\frac 14$ ?

To show that $r=\frac 16$ is the actual radius of convergence, we just need to exhibit one value on the border of the convergence disk that makes the series divergent.

Note that it doesn't prevent the series to converge for other values of $z$ in the concerned ring, but the radius of convergence is defined as the larger $r$ such that the series for $|z|<r$ is guaranteed to converge.

So we try with $z=\frac 16$ and the since the series is with all positive terms we can split it to $\displaystyle \sum\limits_{k=0}^{\infty} \left(\frac 46\right)^{2k+1}+\sum\limits_{k=0}^{\infty} \left(\frac 66\right)^{2k}$

The first series is convergence since it is a geometric series with $\left(\frac 46\right)^2<1$ and the second one is trivially divergent since its terms are all equal to $1\not\to0$.

Thus the series for $z=\frac 16$ is divergent, and it fixes the radius of convergence to its lower bound.

Answer 2

Yes, $e^{i\pi n}=(-1)^n$. Therefore, your series is$$6+4z+6^2z^2+4^3z^3+\cdots$$And, yes, you can use the root test here:$$\limsup_n\sqrt[n]{\left\lvert z^n\bigl(5+(-1)^n\bigr)^n\right\rvert}=6\lvert z\rvert.$$Therefore, the radius of convergence is $\frac16$.