Showing this sequence converges. Given a sequence $x_n=\left(\dfrac{2n^3+n}{n^3} \right)+ i\left(\dfrac{3n}{n+1}\right)$, how would I show it converges? How would I choose $N$?
I did the following.
Given $\epsilon >0,$ choose $N>[?]$. Then for $n>N$
\begin{align}
\left \lvert x_n-(2+3i) \right \rvert & = \left \lvert \frac{2n^3+n}{n^3}-2 \right \rvert+ \lvert \frac{3n}{n+1}-3 \rvert\\
& = \left \lvert \frac{1}{n^2} \right \rvert+ \left \lvert -\frac{3}{n+1} \right \rvert\\
& =\frac{1}{n^2}+\frac{3}{n+1} \\
& = \frac{n+1+3n^2}{n^2+n^3} <?<\epsilon
\end{align}
Hence $x_n \rightarrow 2+3i$.

You reached the point where you want to prove $\,\displaystyle{\frac{3n^2+n+1}{n^3+n^2}<\epsilon}\,$ , but  $$\frac{3n^2+n+1}{n^3+n^2}\leq \frac{5n^2}{n^3}\leq \frac{5}{n}$$ so $\,\displaystyle{\frac{5}{n}<\epsilon \Longrightarrow n>\frac{5}{\epsilon}}\,$ , and thus it is enough to choose $\displaystyle{\,N_\epsilon:=\left[\frac{5}{\epsilon}\right] + 1}\,$ to have that
for any $\,n>N_\epsilon\,$ we have the wanted inequality (with [x] = the integer part of x)
