I have the following power series:
$$ \sum_{n = 1}^{\infty} (1 + \frac{1}{n^{1/n}} )^{n} \sin^{2n}(\frac{n \pi}{3}) z^{n} $$
The question asks me to find the radius of convergence.
Here is my attempt:
I consider the ratio test with $a_n =(1 + \frac{1}{n^{1/n}} )^{n} \sin^{2n}(\frac{n \pi}{3}) z^{n}$
$$ \frac{a_{n+1}}{a_n} = \frac{(1 + \frac{1}{(n+1)^{1/(n+1)}} )^{n+1} \sin^{2n+2}(\frac{(n+1) \pi}{3}) z^{n+1}}{(1 + \frac{1}{n^{1/n}} )^{n} \sin^{2n}(\frac{n \pi}{3}) z^{n}} $$
I tried to simplify this expression, but it seems to be complicated and unapproachable. I am stuck this way. Is there another way to solve this problem?
Denoting $ \left(\forall n\in\mathbb{N}\right),\ b_{n}=\left(1+\frac{1}{\sqrt[n]{n}}\right)^{n} $, and $ \left(\forall n\in\mathbb{N}\right),\ c_{n}=b_{n}\left(\frac{3}{4}\right)^{n} $, and $ \left(\forall n\in\mathbb{N}\right),\ a_{n}=b_{n}\sin^{2n}{\left(\frac{n\pi}{3}\right)} $. $ R_{a} $ is the radius of convergence of $ \sum\limits_{n\geq 0}{a_{n}z^{n}} $, and $ R_{b} $ the radius of convergence of $ \sum\limits_{n\geq 0}{b_{n}z^{n}} \cdot $
Note that $ \left(\forall n\in\mathbb{N}\right),\ \left|\sin{\left(\frac{n\pi}{3}\right)}\right|\leq\frac{\sqrt{3}}{2} $, hence $ \left|a_{n}\right|\leq b_{n}\left(\frac{\sqrt{3}}{2}\right)^{2n} $, and thus $ R_{c}\leq R_{a} \cdot $
Since $ c_{n}\underset{n\to +\infty}{\sim}2^{n}\left(\frac{3}{4}\right)^{n} $, we get that $ R_{c}=\frac{2}{3} $, and thus $ R_{a}\geq \frac{2}{3} \cdot $
Setting $ z=\frac{2}{3} $, we know $ \left(1+\frac{1}{\sqrt[n]{n}}\right)^{n}\sin^{2n}{\left(\frac{n\pi}{3}\right)}\left(\frac{2}{3}\right)^{n}\underset{n\to +\infty}{\sim}\left(\frac{2}{\sqrt{3}}\sin{\left(\frac{n\pi}{3}\right)}\right)^{2n} $, which does not approache $ 0 $ when $ n $ goes to $ +\infty $, which means the series diverges and thus $ R_{a}\leq\frac{2}{3} $
Conclusion : $$ R_{a}=\frac{2}{3} $$