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 $x_n \rightarrow 2+3i$.

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.
