The definition for convergence: The sequence $\{x_n\}$ converges to $L$ where $L\in \mathbb{R}$ provided that for every $\epsilon > 0$ there exists a corresponding integer $N\in \mathbb{N}$ such that $n \geq N$ $\Rightarrow$ $|x_n-L| < \epsilon$.

The sequence given is $$\left\{\frac{2n}{3n+2}\right\}.$$

So far for my proof I have:

Let $\epsilon >0$ be given to us. We must show there is a number N such that $n \geq N\Rightarrow |\frac{2n}{3n+2} - \frac{2}{3}| < \epsilon$.

Then for scratch work I have $|2n/(3n+2) - 2/3| = |6n-6n-4/(9n+6)| = \\|(-4/(9n+6)| = 4/9n+6 \\< 4/9n = 4/9 * 1/n < 1/n <\epsilon $

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    $\begingroup$ The limit is not 7/10. Perhaps plug in a few large numbers to see what fraction it does approach? $\endgroup$
    – ndhanson3
    Mar 3 '21 at 18:13
  • $\begingroup$ ndhanson3- After looking at some larger numbers, it looks like it actually approaching 2/3. $\endgroup$
    – Yogibear
    Mar 3 '21 at 18:20
  • $\begingroup$ @Yogibear It's a paradox, tho. Ho do we know that the sequence converges to $2/3$? $$\frac{2n}{3n+2}=\frac{2n}{n\left(3+\frac{2}{n}\right)}\to 2/3 \text{ as }n\to\infty$$ $\endgroup$
    – Raffaele
    Mar 3 '21 at 18:26

The limit is $\frac23$, not $\frac7{10}$. Note that$$\frac{2n}{3n+2}-\frac23=-\frac4{9n+6}$$and that therefore$$\left|\frac{2n}{3n+2}-\frac23\right|=\frac4{9n+6.}$$So, given $\varepsilon>0$, take $N\in\Bbb N$ such that $N>\frac4{9\varepsilon}$ and then, if $n\geqslant N$,\begin{align}\frac4{9n+6}&<\frac4{9n}\\&\leqslant\frac4{9N}\\&<\varepsilon.\end{align}


Perhaps if you write the nth term of the sequence as $\frac{(2/3)(3n+2)}{3n+2}-\frac{4/3}{3n+2}=\frac{2}{3}-\frac{4/3}{3n+2} $ the limit should become clear and then you just need to find $n$ large enough so that $\frac{4/3}{3n+2}<\epsilon $.


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