Use the definition of convergence to prove that the sequence $\left\{\frac{2n}{3n+2}\right\}$ converges.

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 $$|2n/(3n+2) - 2/3| = |6n-6n-4/(9n+6)| = \\|(-4/(9n+6)| = 4/9n+6 \\< 4/9n = 4/9 * 1/n < 1/n <\epsilon$$

• The limit is not 7/10. Perhaps plug in a few large numbers to see what fraction it does approach? Mar 3 '21 at 18:13
• ndhanson3- After looking at some larger numbers, it looks like it actually approaching 2/3. Mar 3 '21 at 18:20
• @Yogibear It's a paradox, tho. Ho do we know that the sequence converges to $2/3$? $$\frac{2n}{3n+2}=\frac{2n}{n\left(3+\frac{2}{n}\right)}\to 2/3 \text{ as }n\to\infty$$ 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 $$\frac{4/3}{3n+2}<\epsilon$$.