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 $x_n \rightarrow 2+3i$.
 A: 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)
A: The sequence, as you have guessed right, converges to $2+3i$. Your argument is also almost fine. You need to clarify which norm you are using to prove convergence.  Usually one uses the Euclidean norm, in which case, you get 
\begin{align}
\left \lVert x_n - (2+3i) \right \rVert_2 & = \left \lVert \left( \frac{2n^3+n}{n^3} + \frac{3n}{n+1} i  \right) - (2+3i) \right \rVert_2\\
& = \left \lVert \left( \frac{2n^3+n-2n^3}{n^3} \right) + i \left( \frac{3n-3(n+1)}{n+1} \right)\right \rVert_2\\
& = \left \lVert \left( \frac{1}{n^2} \right) - i \left( \frac{3}{n+1} \right)\right \rVert_2\\
& = \sqrt{\left( \frac1{n^2}\right)^2 + \left(\frac{3}{n+1} \right)^2}
\end{align}
Note that for $n \in \mathbb{Z}^+$, we have $\dfrac1{n^2} < \dfrac3{n+1}$. Hence, we get that $$\sqrt{\left( \frac1{n^2}\right)^2 + \left(\frac{3}{n+1} \right)^2} < \sqrt{\left(\frac{3}{n+1} \right)^2 + \left(\frac{3}{n+1} \right)^2} = \frac{3 \sqrt{2}}{n+1}$$
Now given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{3\sqrt{2}}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\dfrac{3 \sqrt{2}}{n+1} < \epsilon.$$
Hence, given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{3\sqrt{2}}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\lVert x_n - \left( 2+3i\right)\rVert_2 < \epsilon$$
Hence, $x_n \rightarrow 2+3i$.

EDIT
You could also use the $1$-norm as you have.
\begin{align}
\left \lVert x_n - (2+3i) \right \rVert_1 & = \left \lVert \left( \frac{2n^3+n}{n^3} + \frac{3n}{n+1} i  \right) - (2+3i) \right \rVert_1\\
& = \left \lVert \left( \frac{2n^3+n-2n^3}{n^3} \right) + i \left( \frac{3n-3(n+1)}{n+1} \right)\right \rVert_1\\
& = \left \lVert \left( \frac{1}{n^2} \right) - i \left( \frac{3}{n+1} \right)\right \rVert_1\\
& = \left \lvert \left( \frac1{n^2} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert
\end{align}
Note that for $n \in \mathbb{Z}^+$, we have $\dfrac1{n^2} < \dfrac3{n+1}$. Hence, we get that $$\left \lvert \left( \frac1{n^2} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert < \left \lvert \left( \frac{3}{n+1} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert = \frac{6}{n+1}$$
Now given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{6}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\dfrac{6}{n+1} < \epsilon.$$
Hence, given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{6}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\lVert x_n - \left( 2+3i\right)\rVert_1 < \epsilon$$
Hence, $x_n \rightarrow 2+3i$.
A: For sequences of complex numbers it occurs that a given sequence of complex numbers $a_{n}=x_{n}+y_{n}i$ converges to some complex number $x+yi$ iff sequence $x_{n}$ converges to $x$ and as one may expect $y_{n}$ converges to $y$. Sequence diverges if at least one of the mentioned conditions  does not hold.
