Convergent vs. Divergent Sequences of same power quotient Alright, so a new topic in math for me, sequences. I'm suppose to prove if the converge or diverge. I'm not to sure how to do this, but i'm gonna put out what I have and see how wrong I am about this.
$${(n-1)(3n+1)^3\over (n-2)^4}$$
So I expanded my equation out and got $${27n^4-18n^2-8n-1\over n^4-8n^3+24n^2+16n+16}$$
So if I took the limit it would be 27 cause of having the same power on both denominator and numerator. However this is not a proof of it converging. How would I use an arbitray epsilon to proves this? My understanding is to use my equation -1. and find an epislon from there, but I can't seem to do it and the book isn't of much help
EDIT: Made a mistake in the writing the equation
$$n*{28n^3+8n^2-36n-24+}{-17\over n }*{1\over n^4-8n^3+24n^2+16n+16 }$$ is what I got by subtracting my equation by 1
 A: A better approach that would still allow $\epsilon$-proofs would be to use upper and lower bounds that are easier to work with and then apply the squeeze theorem. Remember that in order to make a positive fraction larger we can either increase the numerator or decrease the denominator.
For instance, 
$$\frac{(n-1)(3n)^3}{n^4} <\frac{(n-1)(3n+1)^3}{(n-2)^4} < \frac{(n)(3n+1)^3}{(n-2)^4}$$
Reasoning the upcoming step:
Now I'm not a big fan of terms like $(n-1)$ in the above expression, because I know it will make an $\epsilon$ proof a bit messy. We get the same problem with $3n+1$; we could bound it above by $3n+3=3(n+1)$, but then we have to deal with $(n+1)$. I want these expressions in terms of a power of $n$ and maybe some constant multipliers as well. With that in mind:
Fix some $\eta>0$, then for $n$ large enough ($n> N_1$), $$n-1 > n-\eta\cdot n = n(1-\eta)$$
also for $n$ large enough ($n>N_2$),
$$3n+1 < 3n+n\cdot 3\eta = 3n(1+\eta)$$
and finally for $n$ large enough ($n>N_3$),
$$n-2 > n-\eta\cdot n = n(1-\eta)$$
So if $n>\max\{N_1,N_2,N_3\}$,
$$\frac{\big(n(1-\eta)\big)(3n)^3}{n^4} <\frac{(n-1)(3n+1)^3}{(n-2)^4} < \frac{(n)\big(3n(1+\eta)\big)^3}{\big(n(1-\eta)\big)^4}$$
so the squeeze theorem gives
$$27(1-\eta) \lim_{n\to\infty} \frac{n^4}{n^4} \leq \lim_{n\to\infty}\frac{(n-1)(3n+1)^3}{(n-2)^4} \leq 27 \frac{(1+\eta)^3}{(1-\eta)^4} \lim_{n\to\infty} \frac{n^4}{n^4}$$
and an utterly trivial $\epsilon$ proof (that I'll leave to you) shows that $\lim_{n\to\infty} \frac{n^4}{n^4} = 1$. Hence
$$27(1-\eta) \leq \lim_{n\to\infty}\frac{(n-1)(3n+1)^3}{(n-2)^4} \leq 27 \frac{(1+\eta)^3}{(1-\eta)^4}$$
But $\eta>0$ was arbitrary, so letting $\eta \to 0^+$ gives
$$\lim_{n\to\infty}\frac{(n-1)(3n+1)^3}{(n-2)^4} = 27$$
