In $(\mathbb C,{\Vert .\Vert)}$, I have a sequence $x_n=\left(\frac{2n^3+n}{n^3}+\frac{3in}{n+1}\right).$
Learning from my previous question, I found the limit of $x_n$ to be $x=(2+3i)$. But how would this sequence converges?
I worked from $x_n=\left|\left(\frac{2n^3+n}{n^3}+\frac{3in}{n+1}\right)-(2+3i)\right|$
$x_n=\left|\frac{n+n^2+3in^3}{n^4+n^3}\right| < \left|\frac{n+n^2+3in^3}{n^3}\right|$
But how would I continue from there?