real analysis - Converge Test on the series $sum limits_{n=0}^{infty} left(frac{2n+n^3}{3-4n}right)^n$

I want to show, that $a:=\sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n$ is not converging, because $\lim \limits_{n \to \infty}(a)\neq 0 \; (*)$. Therefore, the series can't be absolute converge too.

Firstly, I try to simplify the term. After that I want to find the limit.

Unfortunately, I can't seem to find any good equation with that I can clearly show $(*)$.
$$\begin{align} \sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n&=\left (\dfrac{\not{n}\cdot (2+n^2)}{\not n \cdot (\frac{3}{n}-4)}\right)^n\\ &=\left(\dfrac{2+n^2}{\frac 3n-4}\right)^n\\ &= \cdots \end{align}$$

How to go on?

Answer

Recall that

$$\sum_0^\infty a_n <\infty \implies a_n \to 0$$

therefore if $a_n \not \to 0$ the series can’t converge.

In that case for $n\ge 3$ we have

$$\left|\dfrac{2n+n^3}{3-4n}\right|=\dfrac{2n+n^3}{4n-3}>\dfrac{n^3}{4n}=\frac{n^2}4\ge2$$

and then

$$|a_n|=\left|\dfrac{2n+n^3}{3-4n}\right|^n\ge 2^n$$

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