I was given a task which I solved but now I am not sure if my way is correct, so I would be thankful if someone would verify my solution and maybe give me some tips for improvement.
I have to find an $N\in\mathbb{N}$ for which $a_N-\frac{2}{3}<0,01$ holds for the first time. Here $a_n$ is defined like the following: $a_n:=\frac{2n+6}{3n+8}\in\mathbb{R}$ with $n\in\mathbb{N}$. I did this with equivalent transformations of the inequality and came to the solution that the inequality holds the first time for $N=20$.
Now I have to investigate if the inequality $\lvert a_n-\frac{2}{3}\rvert<0,01$ holds for every $n\geq N$.
This part is giving more points than the first part, which confuses me a bit. Because my idea now was that I only have to show that $a_n-\frac{2}{3}\geq0$ holds. Then I can remove the absolute and would just use the same equivalent transformations than before and I would come to the solution that $\lvert a_n-\frac{2}{3}\rvert<0,01$ indeed holds for every $n\geq N$.
To show that $a_n-\frac{2}{3}\geq0$ I just did the following:
$$a_n-\frac{2}{3}=\frac{2n+6}{3n+8}-\frac{2}{3}=\frac{2}{3(3n+8)}$$ Because $n\in\mathbb{N}$ this is obviously greater than $0$.
Now I can remove the absolute and just do the same as before. Is this everything I have to do or am I missing something?
Whole Solution
Since @CalvinLin pointed out that it's best if I write down my whole solution here it is:
First I have to find an $N\in\mathbb{N}$ for which $a_N-\frac{2}{3}<0,01$ holds the first time. I did this with transformations of the inequality:
$$\begin{align*} & &a_N-\frac{2}{3}&<0,01 & &|\, 0,01=\frac{1}{100}\\ &\Leftrightarrow &\frac{2N+6}{3N+8}-\frac{2}{3}&<\frac{1}{100} & &|\, -\frac{1}{100} \\ &\Leftrightarrow &\frac{2N+6}{3N+8}-\frac{203}{300}&<0 \\ &\Leftrightarrow &\frac{300(2N+6)}{300(3N+8)}-\frac{203(3N+8)}{300(3N+8)}&<0 & &|\,300(3N+8)\neq0, \cdot300(3N+8) \\ &\Leftrightarrow &300(2N+6)-203(3N+8)&<0 \\ &\Leftrightarrow &(600N+1800)-(609N+1624)&<0 \\ &\Leftrightarrow &600N+1800-609N-1624&<0 \\ &\Leftrightarrow &-9N+176&<0 & &|\, -176\\ &\Leftrightarrow &-9N&<-176 & &|\, :(-9)\\ &\Leftrightarrow &N&>\frac{176}{9} \end{align*}$$ We found that the inequation holds the first time for $N=\lceil\frac{176}{9}\rceil=20$.
Now I have to investigate if the inequality $\lvert a_n-\frac{2}{3}\rvert<0,01$ holds for every $n\geq N$.
The following holds: $a_n-\frac{2}{3}=\frac{2n+6}{3n+8}-\frac{2}{3}=\frac{2}{3(3n+8)}$. Because $n\in\mathbb{N}$, $\frac{2}{3(3n+8)}>0\Leftrightarrow a_n-\frac{2}{3}>0$.
Hence $a_n-\frac{2}{3}>0\Rightarrow\lvert a_n-\frac{2}{3}\rvert=a_n-\frac{2}{3}$
From this follows: $\lvert a_n-\frac{2}{3}\rvert<0,01\Leftrightarrow a_n-\frac{2}{3}<0,01$. We are now solving the inequation $a_n-\frac{2}{3}<0,01$ like above and find that $\lvert a_n-\frac{2}{3}\rvert<0,01$ holds for every $n\geq\lceil\frac{176}{9}\rceil$. Because $N=\lceil\frac{176}{9}\rceil$ the inequality $\lvert a_n-\frac{2}{3}\rvert<0,01$ indeed holds for every $n\geq N$.
We wish to find $n$ such that $$\left|a_n - \frac{2}{3}\right| < \frac{1}{100}$$ Since $a_n = \frac{2n + 6}{3n + 8}$, we obtain \begin{align*} \left|\frac{2n + 6}{3n + 8} - \frac{2}{3}\right| & < \frac{1}{100}\\ \left|\frac{6n + 18}{3(3n + 8)} - \frac{6n + 16}{3(3n + 8)}\right| & < \frac{1}{100}\\ \left|\frac{2}{3(3n + 8)}\right| & < \frac{1}{100} \end{align*} If $0 < a < b$, then $\frac{1}{a} > \frac{1}{b}$, which can be seen by multiplying both sides of the inequality $a < b$ by $\frac{1}{ab} > 0$. Hence, \begin{align*} \left|\frac{3(3n + 8)}{2}\right| & > 100\\ |3(3n + 8)| & > 200\\ |9n + 24| & > 200 \end{align*} Since $n \in \mathbb{N}$, $9n + 24 > 0 \implies |9n + 24| = 9n + 24$.
\begin{align*} 9n + 24 & > 200\\ 9n & > 176\\ n & > \frac{176}{9} \end{align*} Hence, the distance of $a_n$ from $2/3$ is less than $1/100$ whenever $n > 176/9$. The smallest such $n$ is $$N = \left\lceil \frac{176}{9} \right\rceil = 20$$ as you found.