1
$\begingroup$

Question: Prove that the sequence $\{a_n\}_{n=1}^\infty$ defined by $a_n=\frac{2n}{3n+1}$ converges to $\frac{2}{3}$ using the $\varepsilon$-$N$ definition of a limit.


I'm still a beginner when it comes to these types of questions. I've had a go at it but I'm pretty sure this is nonsense. Would appreciate some feedback and suggestions. Thanks!


Solution: Fix $\varepsilon>0$. We need to find $N\in\mathbb{N}$ such that $n>N \implies|a_n-\frac{2}{3}|<\varepsilon.$

We have $|a_n-\frac{2}{3}| < \varepsilon.$

$\iff |\frac{2n}{3n+1}-\frac{2}{3}| < \varepsilon$

$\iff |\frac{2n}{3n+1}-\frac{2}{3}| < \varepsilon$

Now $2n\ge2$ for all $n\ge1$ and $3n+1\ge3n$ for all $n$, so we have

$\iff |\frac{2}{3n}-\frac{2}{3}| < \varepsilon$

$\iff |\frac{2}{3}(\frac{1}{n}-1)| < \varepsilon$

The sequence is positive for all $n$ so the absolute values are redundant

$\iff \frac{2}{3}(\frac{1}{n}-1) < \varepsilon$

$\iff \frac{1}{n}-1 < \frac{3\varepsilon}{2}$

$\iff \frac{1}{n} < \frac{3\varepsilon}{2}+1$

$\iff \frac{1}{n} < \frac{3\varepsilon+2}{2}$

$\iff n> \frac{2}{3\varepsilon+2}$

So choose any $N>\frac{2}{3\varepsilon+2}$ and the definition is satisfied. Q.E.D.

$\endgroup$
2
  • $\begingroup$ The proof looks right except for the removing of the absolute value. It is necessary, since $1/n-1$ is negative for all $n\geq 2$. $\endgroup$
    – Dog_69
    Apr 5, 2018 at 10:36
  • 3
    $\begingroup$ it's not right to jump from $\frac{2n}{3n+1}$ to $\frac{2}{3n}$ $\endgroup$
    – Vasili
    Apr 5, 2018 at 10:37

3 Answers 3

4
$\begingroup$

You're certainly making this complicated. And the following step is wrong, and I gave up on trying to figure out why you thought it was true:

$\iff |\frac{2n}{3n+1}-\frac{2}{3}| < \varepsilon$

Now $2n\ge2$ for all $n\ge1$ and $3n+1\ge3n$ for all $n$, so we have

$\iff |\frac{2}{3n}-\frac{2}{3}| < \varepsilon$

Remember that the way to subtract fractions is by using a common denominator: $$ \left| \frac{2n}{3n+1} - \frac 2 3 \right| = \left| \frac{3\cdot2n}{3(3n+1)} - \frac{2(3n+1)}{3(3n+1)} \right| = \left| \frac{6n - (6n+2)}{3(3n+1)} \right| = \frac 2 {3(3n+1)} $$

The problem now is how to make $n$ so big that that last fraction is less than $\varepsilon.$

$\endgroup$
1
  • $\begingroup$ Thanks for the help. Using this I obtained $N>\frac{2}{9\epsilon}$, is that correct? $\endgroup$
    – Sonjov
    Apr 5, 2018 at 11:58
2
$\begingroup$

HINT: it is $$\frac{2n}{3n+1}-\frac{2}{3}=\frac{-2}{3(3n+1)}$$ and $$\left|\frac{-2}{3(3n+1)}\right|=\frac{2}{3(3n+1)}$$

$\endgroup$
0
$\begingroup$

Complete the $\epsilon-n_0$ argument:

We have

$|\dfrac{2n}{3n+1} -\dfrac{2}{3}|=\dfrac{2}{3(3n+1)} \lt$

$\dfrac{2}{3(3n)}\lt \dfrac {2}{n}.$

Let $\epsilon >0$ be given.

Archimedes :

There is a $n_0 \in\mathbb{Z^+}$ with

$n_0 \gt 2/\epsilon.$

For $n \ge n_0:$

$|\dfrac{2n}{3n+1} -\dfrac{2}{3}|=$

$\lt \dfrac{2}{n} \le \dfrac{2}{n_0} \lt \epsilon$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .