# Convergence of complex power series $\sum_n \dfrac{(-4+4\sqrt3i)^n}{n} \dfrac{1}{(z+3)^{6n}}$

The problem

Find where does the following series converge and where does it converge absolutely $$\sum_n \dfrac{(-4+4\sqrt3i)^n}{n} \dfrac{1}{(z+3)^{6n}}$$.

My attempt. I first took the following substitution $$w=\dfrac{1}{(z+3)^6}$$, so I'm left with the following power series $$\sum_n \dfrac{(-4+4\sqrt3i)^n}{n} w^n$$ which I can find the radius of convegernce in terms of $$|w|$$ by using the Cauchy-Hadamard formula. Now I'm left to see if there's convergence on the border of that region. For that I'm having trouble deciding. I can't use Dirichlet criterion because the main sequence is not decreasing nor I can use Dedekind criterion. So I'm led to believe that maybe the series doesn't converge on the border, but I can't find a suitable series to compare to.

For what it's worth, if I made the problem a little bit easier, for example by taking $$w=\dfrac{1}{(z+3)}$$ I can find the radius of convergence of $$\sum_n w^n$$, but I can't relate that with $$\sum_n a_n w^n$$ if $$a_n$$ doesn't converge to 0.

Any hints would be apreciated.

• If a question similar like this has been asked, feel free to share the link. It's not really that easy to find questions about similar series to these one. – Leo Lerena Oct 18 '18 at 21:15
• there is a similar question here math.stackexchange.com/questions/135056/…, btw you can write you series as $\sum a_n x^n$ where $a_n =1/n$ and $x^n =$ all others terms – ALG Oct 18 '18 at 21:22
• just a note: this is not a power series – zhw. Oct 18 '18 at 23:05

## 2 Answers

First note that $$-4+i4\sqrt 3=8e^{i2\pi/3}$$. Hence, we see that

\begin{align}\left|\frac{\left(-4+i4\sqrt 3\right)^n}{(z+3)^{6n}}\right|&=\left|\frac{8e^{i2\pi/3}}{(z+3)^6}\right|^n\\\\ &=\left(\frac{8}{|z+3|^6}\right)^n \end{align}

Finally, taking the $$n$$'th root reveal

\begin{align} \lim_{n\to\infty}\sqrt[n]{\left|\frac{\left(-4+i4\sqrt 3\right)^n}{n(z+3)^{6n}}\right|}&=\frac{8}{|z+3|^6}\lim_{n\to\infty}n^{-1/n}\\\\ &=\frac{8}{|z+3|^6}<1 \end{align}

when $$|z+3|>8^{1/6}$$. So, the series converges for $$|z+3|>8^{1/6}$$.

What Happens on the Circle $$|z+3|=8^{1/6}$$? Well, if $$(z+3)=8^{1/6}e^{i\theta}$$, then

\begin{align} \sum_{n=1}^\infty \frac{(-4+i4\sqrt3)}{n(z+3)^{6n}}&=\sum_{n=1}^\infty \frac{e^{i2n\pi/3}}{ne^{i6n\theta}}\\\\ &=\sum_{n=1}^\infty \frac{e^{i(6\theta +2\pi/3)n}}{n} \end{align}

Now, apply Dirichlet's Test to find that range of values of $$\theta$$ for which there exists a number $$N$$ such that

$$\left|\sum_{n=1}^N e^{i(6\theta+2\pi/3)n}\right|

for all $$N$$.

Can you finish now?

• Thanks for the answer. If $\theta$ happens to be such that $6 \theta + 2\pi / 3 = 0$ then it doesn't converge. Otherwise the series converges because this is a geometric series. Applying Dirichlet, as the sequence $1/n$ is decreasing and tends to $0$, I can conclude that the series converges on the boundary except for one point and that it converges absolutely for $|z+3|>8^{1/6}$ – Leo Lerena Oct 18 '18 at 22:09
• If $6\theta +2\pi/3=2k\pi$ for $k\in\mathbb{Z}$, then the series diverges. – Mark Viola Oct 18 '18 at 22:43
• If $6\theta +2\pi/3=2k\pi$ for $k\in\mathbb{Z}$, then the series diverges. – Mark Viola Oct 18 '18 at 22:43

Note that$$\sqrt[n]{\left\lvert\dfrac{(-4+4\sqrt3i)^n}{n} \dfrac{1}{(z+3)^{6n}}\right\rvert}=\frac{\left\lvert-4+4\sqrt3i\right\rvert}{\sqrt[n]n\lvert z+3\rvert^6}=\frac8{\sqrt[n]n\lvert z+3\rvert^6}\to\frac8{\lvert z+3\rvert^6}.$$Therefore, the series converges if $$\lvert z+6\rvert>\sqrt2$$ and diverges if $$\lvert z+3\rvert<\sqrt2$$.

If $$\lvert z+3\rvert=\sqrt2$$, there are two possibilities:

• $$(z+3)^6=-4+4\sqrt3i$$: then your series is the harmonic series, which diverges.
• $$\lvert z+3\rvert=\sqrt2$$, but $$(z+3)^6\neq-4+4\sqrt3i$$: then your series converges, by Dirichlet's test.