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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?

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    Yes. OK. Then it is better not to delete the $n^4$ term, but the $n^3$ term.2012-02-20

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You can make life simpler if you avail yourself of the fact that a sequence of complex numbers converges to $z$ iff its real part converges to $\Re(z)$ and its imaginary part converges to $\Im(z)$.

This simplifies the algebra. For the real part, you have $\left|{2n^3 + n\over n^3} - 2\right| = {1\over n^2} \rightarrow 0\; {\rm as}\; n\to\infty. $

For the imaginary part, you have $\left|{3n\over n+1} - 3\right| = {3\over n+1} \to 0 \;{\rm as}\; n\to\infty. $