Trouble understanding the $ε$-$N$ proof for limit of sequence

Question

I'm currently stuck on a math problem proposed by my teacher.

For $\varepsilon=0,001$, find $N=N_0$ such that $|a_n-L|\lt \varepsilon$ if $n\geqslant N$

$$a_n = \frac{(n+1)}{3n-1}\, {\rm and} \, L=1/3$$

According to him, the answer could be any $N_0$ greater than $\left[\frac{2}{3\varepsilon}\right]+1$. I have an idea of how the proof works, but I can't figure out how he came up with that specific value for N$_0$. And why the final value has a $+1$.

Answer 1

We have that

$$\left|\frac{n+1}{3n-1}-\frac13\right|<\epsilon \iff \frac{4}{3(3n-1)}<\epsilon. $$

Now

$$\frac{4}{3(3n-1)}<\epsilon\iff 9n-3>\frac{4}{\epsilon}\iff 9n>\frac{4}{\epsilon}+3.$$ So we need

$$n>\frac{4}{9\epsilon}+\frac13.$$ So it is enough to have

$$n>\left[\frac{4}{9\epsilon}\right]+\left[\frac13\right]=\left[\frac{4}{9\epsilon}\right]+1.$$

Answer 2

We have $$ a_n - L = \frac{n+1}{3n-1}- \frac{1}{3} = \frac{3(n+1)-(3n-1)}{3(3n-1)} = \frac{4}{3(3n-1)}. $$ So we want $$ \frac{4}{3(3n-1)} < \epsilon \quad\Leftrightarrow\quad 3n-1 > \frac{4}{3\epsilon} \quad \Leftrightarrow \quad n > \frac{4}{9\epsilon}+\frac{1}{3}, $$ so we can take $$ n_0 = \left[ \frac{4}{9\epsilon}\right] + 1. $$ We have $+1$ here to be on the safe side, since we want the previous inequality to be strict.

In particular, if $\epsilon = 0.001 = 10^{-3}$ this becomes $$ n_0 = \left[ \frac{4000}{9}\right]+1 = \left[ \frac{4000}{9}\right]+1 = 445. $$