Proof for sequence Let $x_n=(-1)^n\frac{n^2}{3^n}$ Let $x_n=(-1)^n\frac{n^2}{3^n}$
be a real sequence. 
It is claimed that $x_n \rightarrow 0$ for $n\ge4$.
New to real analysis, I am having problems with the correct form for the proof. 
Let me try and I hope for your corrections (ideally with explanations).
1.) Let $\epsilon>0$
2.) There exists a natural number K, such that for any $n: n\ge K$  satisfies $(|x_n-x|)<\epsilon$. We want to find such K, call it $n_k$. 
3.) Here is the ??? In other proofs, the x is replaced with the known convergence limit, and the inequality is solved for n. However, in this case it is not so intuitive. 
3.1) Is it possible to simply state(?): According to the Archemedian Property it must hold that $\frac{1}{n_k}<\epsilon$ therefore $\frac{n^2}{3^n}\le\frac{1}{n_k}<\epsilon $
4.) It follows $\frac{n^2}{3^n}\le\frac{1}{n_k} \Rightarrow n^3\le3^n $ must hold for $n \ge4$ (by testing 64<81)
5.) Lemma: for any $n\in N: n\ge4 $ such that $n^3\le3^n$
6.) By mathematical induction the general case of the lemma is proven for $n\ge4 $. Assuming the statement is true for n=k, then it must also hold for n=k+1. Shown in two steps: a) $3^k\ge3k^2+3k+1$ and b) $3^k\ge k^2$ The first is required in the proof of the second. (excluded here)
It is proven that $3^{k+1}\ge(k+1)^2$ Therefore $\frac{n^2}{3^n} \rightarrow 0$ as $n \rightarrow \infty$ given the lemma. (the denominator is larger than the nominator). 
7.) Then  $\frac{n^2}{3^n}\le\frac{1}{n_k}<\frac1\epsilon $ Thus for any $x \in R$, exists $n\in N$, such that $ n>x$. And for $\frac{1}{n_k}<\epsilon$ by the Archimedian Property  we known such a $n_k$ exists in N. 
8.) Given lemma let E>0, let $n_k>\frac1\epsilon$. Then if $n\ge n_k \Rightarrow |x_n-L|< \epsilon$, where L=0.
Thank you for any corrections. 
 A: 3.1 makes no sense.  $\frac{n^2}{3^n}\le\frac{1}{n_k}<\epsilon$ is what you're trying to prove.  The "Archimedian property" does not seem to help.   
You're on the right track trying to prove that for large $n$ we have $n^3\le 3^n$.  Here is a rewrite of your proof.  We have $4^3=64<3^4=81$.  Now suppose $n^3\le 3^n$; we want to show $(n+1)^3\le 3^{n+1}$, and then our result is proven by induction.  We have $(n+1)^3=n^3+3n^2+3n+1$.  We have $n^3> 3n^2$ because $n\ge 4$, and $n^3\ge 3n+1$ because $n^3>3n$ for $n\ge 4$. This shows that $(n+1)^3\le n^3+n^3+n^3\le 3\cdot 3^n=3^{n+1}$ as desired.  
Once you've got $n^3\le 3^n$, you have $n^2/3^n= n^3/3^n\cdot (1/n)\le 1/n$.  This shows that $|(-1)^nn^3/3^n|=n^3/3^n\le 1/n$ and thus $(-1)^nn^3/3^n\to 0$ as $n\to\infty$ as you wished.
Now, what about epsilons and deltas?  If you want to use them here, I would suggest proving a more general result which you can use in the last step just given:  if $f$ is a real function defined on the natural numbers and $|f(n)|\le 1/n$ for large enough $n$, then $\lim f(n)=0$.  That limit by definition requires that for any given $\epsilon>0$, we can choose $n$ such that $|f(n')-0|<\epsilon$ for $n'\ge n$.  In our case, given $\epsilon>0$, we can just choose $n$ such that $1/n<\epsilon$, and then we'll have $|f(n')-0|=|f(n')|\le 1/n'<\epsilon$ for $n'\ge n$ as desired.  
