Q:

Accumulation points of a recursive sequence $a_n=\cos\left(\frac{2\pi n}{3}\right)\sqrt[n]{1+2^n+(-3)^n}$

Find all the accumulation points of the sequence $(a_n)_{n=1}^{\infty}$ defined recursively:

$$a_n=\cos\left(\frac{2\pi n}{3}\right)\sqrt[n]{1+2^n+(-3)^n}$$

My attempt:
$$\cos\left(\frac{2\pi n}{3}\right)\in\left\{-\frac{1}{2},1\right\}$$
$$\sqrt[n]{1+2^n+(-3)^n}=\sqrt[n]{(-3)^n\left(\left(-\frac{1}{3}\right)^n+\left(-\frac{2}{3}\right)^n+1\right)}=-3\sqrt[n]{\left(-\frac{1}{3}\right)^n+\left(-\frac{2}{3}\right)^n+1}$$
I tried to manipulate with the above expression and apply the squeeze theorem, but I haven't managed to get anything useful for this problem, except:
$$\sqrt[n]{1+2^n+(-3)^n}<\sqrt[n]{3\cdot3^n}=3\sqrt[n]{3}$$ I also considered artithmetic inequalities and the comparison: $1+2^n\leq(1+2)^n$, unsuccessfully.
I would definitely examine the cases $n\pmod{6}$ or $n\pmod{3}$. How can I move further from this point and what can I do with the $n-\text{th}$ root? I thought I could see how one of the factors behaves since $\cos\left(\frac{2\pi n}{3}\right)$ alternates for $n\in\mathbb N$. Thank you in advance!

A:

When $n$ is even, 
$$\tag1
(1+2^n+(-3)^n)^{1/n}=(1+2^n+3^n)^{1/n}\to3.
$$
when $n$ is odd, 
$$
(1+2^n+(-3)^n)^{1/n}\to-3.
$$
If you consider any subsequence $a_{n_k}$ that contains both even and odd $n_k$ for arbitrarily big $k$, the subsequence will fail to be convergent. So we only get convergent subsequences when the $n_k$ are eventually all even or odd. Thus the possible limits are 
$$
-\tfrac12\times3,-\tfrac12\times(-3),3,-3.
$$
That is, the accumulation points are 
$$
-\frac32,\frac32,3,-3. 
$$

To calculate the limit $(1)$, you have 
\begin{align}
(1+2^{2n}+(-3)^{2n})^{1/2n}&=(1+2^{2n}+3^{2n})^{1/2n}=3(1+3^{-2n}+(2/3)^{2n})^{1/2n}\\ \ \\
&=3\exp\left(\tfrac1{2n}\,\log(1+3^{-2n}+(2/3)^{2n})\right)\\ \ \\
&=3\exp\left(\tfrac{3^{-2n}+(2/3)^{2n}+o((2/3)^{4n})}{2n} \right)\\ \ \\
&\to3.
\end{align}

