Prove that $(4-\frac{1}{n^4})(-\frac{1}{3})^n$ converges to zero using $\epsilon$-$n$ definition

Question

My attempt so far is:

$\forall \epsilon>0$, $\exists N\in\mathbb{N}\;\forall\;n\geq N:$

$$|(4-\frac{1}{n^4})(-\frac{1}{3})^n-0|<\epsilon$$

Since $(-\frac{1}{3})^n<1 \;\forall \,n\in \mathbb{N}$,

$$|(4-\frac{1}{n^4})(-\frac{1}{3})^n|<|(4-\frac{1}{n^4})|$$

$$|(4-\frac{1}{n^4})|=4-\frac{1}{n^4}$$

We choose an $N$ such that $N<\sqrt[4]{\frac{1}{4-\epsilon}}$

$$4-\frac{1}{n^4}<4-\frac{1}{N^4}<\epsilon.$$

Therefore the sequence converges to $0$.

Answer 1

Your proof is not correct. $$4-\frac{1}{n^4}\geq 4-\frac{1}{N^4}$$ if $n \geq N$ and your proof breaks down here.

$4-\frac 1 {n^{4}} <4$ and $\left|\left(-\frac 1 3\right)^{n}\right|<\epsilon /4$ if $3^{n}>\frac 4 {\epsilon}$. Can you finish from here? [Take logarithm].