Convergent vs. Divergent Sequences of same power quotient

Question

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

Answer 1

Claim: We have $$ \lim_{n\to\infty} \frac{(n-1)(3n+1)^3}{(n-2)^4} = 27.$$

Proof: In order to prove this, we must show that for any $\varepsilon > 0$, there exists some $N > 0$ such that $$ \left| \frac{(n-1)(3n+1)^3}{(n-2)^4} - 27 \right| < \varepsilon $$ for any $n \ge N$. So, begin by fixing some $\varepsilon > 0$. Then (with a bit of help from WolframAlpha to expedite the tedious computations)

\begin{align} \left| \frac{(n-1)(3n+1)^3}{(n-2)^4} - 27 \right| &= \left| \frac{(27n^4 - 18n^2 - 8n - 1) - 27(n^4 - 8n^3 + 24n^2 + 16n + 16)}{n^4 - 8 n^3 + 24 n^2 - 32 n + 16} \right| \\ &= \left| \frac{216 n^3 - 666 n^2 + 856 n - 433}{n^4 - 8 n^3 + 24 n^2 - 32 n + 16} \right| \\ &= \frac{1}{n} \left| \frac{216 - 666 \frac{1}{n} + 856 \frac{1}{n^2} - 433\frac{1}{n^3}}{ 1 - 8 \frac{1}{n} + 24 \frac{1}{n^2} - 32 \frac{1}{n^3} + 16 \frac{1}{n^4}} \right|. \end{align} By choosing $n$ large enough, we can ensure that $$ \left| 1 - 8 \frac{1}{n} + 24 \frac{1}{n^2} - 32 \frac{1}{n^3} + 16 \frac{1}{n^4} \right| > \frac{1}{2}; $$ indeed, via the tried-and-true method of guess-and-check (with a little help from our friend WolframAlpha) $n > 13$ seems to get the job done. This implies that \begin{align} \frac{1}{n} \left| \frac{216 - 666 \frac{1}{n} + 856 \frac{1}{n^2} - 433\frac{1}{n^3}}{ 1 - 8 \frac{1}{n} + 24 \frac{1}{n^2} - 32 \frac{1}{n^3} + 16 \frac{1}{n^4}} \right| &< \frac{2}{n} \left| 216 - 666 \frac{1}{n} + 856 \frac{1}{n^2} - 433\frac{1}{n^3} \right| \\ &\le \frac{2}{n} \left( 216 + 666\frac{1}{n} + 856 \frac{1}{n^2} + 433 \frac{1}{n^3} \right). \end{align} In this last inequality, we are using the triangle inequality and the fact that $n > 0$ in order to obtain the result. Again, we can pick $n$ large enough to get a somewhat sloppy estimate. Indeed, if $n > 666$, then $\frac{666}{n} < 1$, $\frac{856}{n^2} < 1$, and $\frac{433}{n^3} < 1$. Thus, with this very sloppy estimate, we have $$ \frac{2}{n} \left( 216 + 666\frac{1}{n} + 856 \frac{1}{n^2} + 433 \frac{1}{n^3} \right) < \frac{2}{n} \left( 216 + 1 + 1 + 1 \right) = \frac{438}{n}. $$ Finally, if $n > 438\frac{1}{\varepsilon}$, we get $$ \frac{438}{n} < \frac{438}{438 \frac{1}{\varepsilon}} = \varepsilon. $$ Therefore if we take $N \ge \max\left\{13, 666, 438\frac{1}{\varepsilon} \right\}$, then for any $n \ge N$, we have $$ \left| \frac{(n-1)(3n+1)^3}{(n-2)^4} - 27 \right| < \varepsilon, $$ which gives the desired result.

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Note that there are better ways of doing this. Typically, you want to prove two results:

Lemma: We have $$ \lim_{n\to\infty} \frac{1}{n} = 0. $$

You should be able to prove this without too much difficulty.

Lemma: Suppose that $\lim_{n\to\infty} a_n = A$ and that $\lim_{n\to\infty} b_n = B$. Then $$ \lim_{n\to\infty} (a_n + b_n) = A+B. $$ Moreover, if $B \ne 0$, then $$ \lim_{n\to\infty} \frac{a_n}{b_n} = \frac{A}{B}. $$

The proof of the first identity is pretty straight-forward. The second is a little more delicate, but also very doable. You should perhaps go through the details of each—they are good exercises. Once these results are established, we have \begin{align} \lim_{n\to\infty} \frac{(n-1)(3n+1)^3}{(n-2)^4} &= \lim_{n\to\infty} \frac{27n^4 - 18n^2 - 8n - 1}{n^4 - 8 n^3 + 24 n^2 - 32 n + 16} \\ &= \lim_{n\to\infty} \frac{27 - \frac{18}{n^2} - \frac{8}{n^3} - \frac{1}{n^4} }{1 - \frac{8}{n} + \frac{24}{n^2} - \frac{32}{n^3} + \frac{16}{n^4} } && \left(\text{multiply by $\frac{1/n^4}{1/n^4} = 1$}\right) \\ &= \frac{\lim_{n\to\infty} \left( 27 - \frac{18}{n^2} - \frac{8}{n^3} - \frac{1}{n^4} \right) }{\lim_{n\to\infty} \left( 1 - \frac{8}{n} + \frac{24}{n^2} - \frac{32}{n^3} + \frac{16}{n^4} \right) }. \end{align} Since $\lim_{n\to\infty} \frac{1}{n}$, most of the terms in the numerator and denominator go to zero. Thus $$ \frac{\lim_{n\to\infty} \left( 27 - \frac{18}{n^2} - \frac{8}{n^3} - \frac{1}{n^4} \right) }{\lim_{n\to\infty} \left( 1 - \frac{8}{n} + \frac{24}{n^2} - \frac{32}{n^3} + \frac{16}{n^4} \right) } = \frac{27 + 0 + 0 + 0}{1 + 0 + 0 + 0 + 0} = 27.$$

Answer 2

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 $$\lim_{n\to\infty}\frac{(n-1)(3n+1)^3}{(n-2)^4} = 27$$