<s>[INST] Prove that \frac{2n}{3n+7} converges (Question about my method of proof) Show that the sequence $<\frac{2n}{3n+7}>$ converges. Find its limit.
This is the work I have currently:

$\displaystyle \lim_{n \to \infty} \frac{2n}{3n+7} = \frac{2}{3}$
Let $\epsilon > 0$, and observe that:
  \begin{align*}
    \left|\frac{2n}{3n+7} - \frac{2}{3}\right| &= 
    \left|\frac{3(2n) - 2(3n + 7)}{3(3n + 7)}\right|\\
    &= \left|\frac{-14}{9n+21}\right|\\
    &= \frac{14}{9n+21}
  \end{align*}
  We want $\displaystyle \frac{14}{9n+21} < \epsilon$. Solving for $\epsilon$ gives us:
  \begin{align*}
    \displaystyle \frac{14}{9n+21} &< \epsilon\\
    14 &< \epsilon(9n + 21)\\
    \frac{14}{\epsilon} &< 9n + 21\\
    \frac{14}{\epsilon} - 21 &< 9n\\
    \frac{14}{9 \cdot \epsilon} - \frac{7}{3} &< n
  \end{align*}
  So, choose $N$ such that $N = \displaystyle \frac{14}{9 \cdot \epsilon} - \frac{7}{3}$. Then for all $n > N$, we have $\displaystyle \left|\frac{2n}{3n+7} - \frac{2}{3}\right| < \epsilon$.

This is the result I came up with for my homework question. I was wondering if when proving convergence the "scratch work" for finding $N$ is included in the proof. (If there is anything else wrong with my proof please let me know)
 [/INST] If you just take $$N=\left\lceil\frac{14}{9\epsilon}-\frac73\right\rceil$$ so that $N$ is an integer, then you'll be set. (It actually doesn't matter whether $N$ is an integer, but some texts do specify that $N$ is an integer in their definitions.)
It isn't necessarily required that the "scratch work" be shown in a proof. You can always just pull it out of your hat, and say that if $n>N,$ then $n>\frac{14}{9\epsilon}-\frac73,$ so $9n>\frac{14}\epsilon-21,$ so....
Basically, you'll just be doing the scratchwork in reverse.
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