Prove that the sequence $a _n = \frac{n}{3n^3-\sin(n)}$ is convergent using the definition of limit

Question

I tried solving it and got:

Suppose $\varepsilon>0$ and let $L=0.$ Then for all $n$ that is an element of $\mathbb{N}$ with $n\geq N$, $$|a _n-0|=\left|\frac{n}{3n^3-\sin(n)}\right|=\frac{n}{3n^3-|\sin(n)|} \leq \frac{n}{3n^3-1}$$

I should somehow algebraically manipulate above to prove that $|a _n-0|<\varepsilon$ by showing that $N=$ something with $\varepsilon$, but can't get my head around to the next steps. Please help

Answer 1

As pointed out by Peter, you're not using the absolute value right. However, it can be removed as $|\sin(n)| \leq 1 \implies 3n^3 - \sin(n) \geq 0$ for $n \geq 2$ (we can ignore $x_1$). Let $N$ be an integer large enough so that $\frac{1}{2N^2} < \varepsilon$. Observe that, since $n \geq 1 \implies n^3 \geq 1$ we have: $$ |a_n| = \frac{n}{3n^3 - \sin(n)} \leq \frac{n}{3n^3 - n^3} = \frac{1}{2n^2} $$ Thus, for $n \geq N$: $$ |a_n| \leq \frac{1}{2n^2} \leq \frac{1}{2N^2} < \varepsilon $$

Answer 2

$n>0; 3n^2 > 3> 1$ so $0 < 3n^2 -1 \le 3n^2 - \sin n \le 3n^2 + 1$ (actually, the last is a strict inequality as $\sin$ of a non-zero integer is never $\pm 1$, but that's not really important to take into account) and so

$0 < \frac n{3n^2 + 1} \le \frac n{3n^2 - \sin n}=a_n < \frac n{3n^2-1}$

So now we can start:

We fix an arbitrarily small $\epsilon$ and want to find an appropriate $N$ so that $n>N$ would imply

$|a_n -0| = |\frac n{3n^2-\sin n}- 0| \le \frac n{3n^2-1}< \epsilon$.

(You might say but we don't need $\frac n{3n^2-1}$, we might have $\frac n{3n^2-\sin n}<\epsilon \le \frac n{3n^2-1}$... well, we are overdoing it. If we show that $\frac n{3n^2-1} < \epsilon$ then it will will follow that $|\frac n{3n^2 -\sin n}-0| \le \frac n{3n^2-1} < \epsilon$. There is nothing wrong with over kill).

So we want $\frac n{3n^2-1}< \epsilon$ so what do we need for $n$ to make that happen?

We need:

$\frac n{3n^2 -1} = \frac 1{3n -\frac 1n} < \epsilon$ or

$\frac 1{\epsilon} < 3n - \frac 1n$.

Now, remember.... there is nothing wrong with overkill....

$n \ge 1$ so $0 < \frac 1n < 1$ so $3n-1 < 3n-\frac 1n < 3n$

We need $\frac 1{\epsilon} < 3n - \frac 1n$ and we can make that happen if we can make

$\frac 1{\epsilon} < 3n - 1 < 3n-\frac 1n$ happen.

So if we have $3n > \frac 1{\epsilon} + 1$ or $n > \frac 1{3\epsilon}+\frac 13$.

So if $N = \frac 1{3\epsilon}+\frac 13$ we are done.

.....

Now, I think one potential aspect of confusion is that it's not often emphasized these types of proofs are all working BACKWORDS. Normally our proves go in the $\implies$ direction we start with a loose hypotheses and get tighter and tighter conclusions. In this proofs we start with tight conclusion $|a_n-0|<\epsilon$ and get a series of looser and looser conditions that allow it to be true.

I think for good practice it's a good idea to work the proof FORWARD to see how it works.

Let $\epsilon > 0$ be any arbitrarily small value.

If $N = \frac 1{3\epsilon} + \frac 13$ and $n \in \mathbb N$ and $n> \frac 1{3\epsilon} + \frac 13$ then

$3n- 1 > \frac 1{\epsilon}$

and $\epsilon > \frac 1{3n -1}$. Now $n\ge 1$ so

$\frac 1{3n- 1} > \frac 1{3n - \frac 1n} = \frac n{3n^2 - 1}$. Now $1\ge \sin n$ so

$\frac n{3n^2-1} \ge \frac n{3n^2-\sin n}$. Now $n> 0; 3n^2 > 1 \ge \sin n$ so $\frac n{2n^2-\sin n}$ is positive so

$\frac n{3n^2-\sin n} = |\frac n{3n^2-\sin n}|=|\frac n{3n^2-\sin n}-0|=|a_n -0$.

So we conclude:

If $n > \frac 1{\epsilon} + \frac 13$ then

$\epsilon > |a_n-0|$ which is the definition of $\lim_{n\to \infty} a_n = 0$.