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\noindent Recall $\mathbb{N}=\{1,2,3,\ldots\}$. Definition (directly from book) of convergence of the sequence $(a_n)$ to the real number $b$, i.e. the formal meaning of ``$\displaystyle\lim_{n \rightarrow \infty} a_n = b$":
$$\forall \epsilon>0 \exists N \in \mathbb{N}, \hspace{2pt} \mathrm{such \hspace{2pt} that \hspace{2pt}} n>N \rightarrow (|a_n-b|<\epsilon).$$
In class, I wrote a slightly different, but equivalent thing (I was not careful to mention that $N \in \mathbb{N}$, but our author is similarly not careful to mention that $n \in \mathbb{N}$ in his definition, although that should be clear from its use in writing the sequence ``$a_n$"):
$$\forall \epsilon > 0 \exists N > 0 \forall n>N (|a_n-b|<\epsilon).$$
\begin{example}
Prove that $\displaystyle\lim_{n \rightarrow \infty} \dfrac{n}{n+1}=1$. 
\end{example}
\begin{proof}
Let $\epsilon>0$, let $N>\dfrac{1}{\epsilon}-1$, and let $n>N$ (\textit{note: if $\epsilon > 1$, then the requirement that $N \in \mathbb{N}$ means we can take, say, $N=1$}). For such $n$, we have
\[ n+1 > N+1 > \left( \dfrac{1}{\epsilon}-1 \right) + 1 = \dfrac{1}{\epsilon}, \]
and thus $\dfrac{1}{n+1} < \epsilon$. Therefore, calculate
\[
\left| \dfrac{n}{n+1} - 1 \right| = \left| \dfrac{1}{1+n} \right| < \epsilon,
\]
completing the proof. $\blacksquare$
\end{proof}

\begin{example}
Prove that $\displaystyle\lim_{n \rightarrow \infty} \dfrac{2n}{3n+5} = \dfrac{2}{3}$.
\end{example}
\begin{proof}
Let $\epsilon>0$, let $N>\dfrac{10}{9\epsilon} - \dfrac{5}{3}$, and let $n>N$. For such $n$, we have
$$3n+5 > 3N+5 > 3 \left( \dfrac{10}{9\epsilon} - \dfrac{5}{3} \right)+5=\dfrac{10}{3\epsilon}.$$
Thus $\dfrac{9n+15}{10} > \dfrac{1}{\epsilon}$ and we have $\dfrac{10}{9n+15} < \epsilon$. Therefore, calculate
$$\begin{array}{ll}
\left| \dfrac{2n}{3n+5} - \dfrac{2}{3} \right| &= \left| \dfrac{6n-(6n+10)}{9n+15} \right| \\
&=\left| \dfrac{-10}{9n+15} \right| \\
&=\dfrac{10}{9n+15} \\
&< \epsilon,
\end{array}$$
completing the proof. $\blacksquare$
\end{proof}


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