Prove that this sequence converges to 27: $$\{\frac{(n-1)(3n+1)^3}{(n-2)^4}\}$$ Here is my proof:
Want to show: $\forall\epsilon>0,\exists N > 0| \forall n \in \mathbb N, n>N \implies|\frac{(n-1)(3n+1)^3}{(n-2)^4}-27|<\epsilon$
$$|\frac{(n-1)(3n+1)^3}{(n-2)^4}-27|$$ $$<|\frac{n(3n+1)^3}{(n-2)^4}-27|$$ $$\text{... do some arithmetic for common denominator and simplification...}$$ $$=|\frac{243n^3-639n^2+865n-423}{n^4-8n^3+24n^2-32n+16}|$$
But then I'm stuck on this step. How can I simplify this so that I can choose a $N$ with respect to $\epsilon$?
You don't necessarily need to simplify, but rather to "majorate", note that:
Find a larger numerator:
1- If $n>0$ $$ 243n^3−639n^2+865n−423<243n^3+865n $$
2-if $n>\sqrt(865)$: $$ n^3>865n $$ Thus: $$ 243n^3−639n^2+865n−423<244n^3 $$
Find a smaller denominator:
For $n>0$: $$ n^4−8n^3+24n^2−32n+16>n^4−8n^3−32n $$
For $n>8$: $$ n^4−8n^3−32n>n^4−9n^3 $$
So: $$ \frac{243n^3−639n^2+865n−423}{n^4−8n^3+24n^2−32n+16}>\frac{244n^3}{n^4−9n^3}=\frac{244}{n−9}<\varepsilon $$
Now it suffices to pick: $$ n>244/\varepsilon+9 $$