For this question, I know that the sequence converges, but I get stuck in the middle of the proof. Here is what I have so far. Can anyone please help me out?
Determine whether the sequence is convergent or divergent.
$$\left\{\frac{(n-1)(3n+1)^3}{(n-2)^4}\right\}$$
$$\lim_{n \to \infty}\left(\frac{(n-1)(3n+1)^3}{(n-2)^4}\right) = 27$$
Let $\epsilon > 0$ be arbitrary. Suppose $n>N$
\begin{align} |a_n-L|=&\; \left|\frac{(n-1)(3n+1)^3}{(n-2)^4}-27\right| =&\; \left|\frac{(n-1)(3n+1)^3-27(n-2)^4}{(n-2)^4}\right| \end{align}
This question is a duplicate. I'll answer again anyway.
The typical trick is to find upper or lower bounds for the terms involved:
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: $$ n>244/\varepsilon+9 $$