Real and complex limits

Show that a complex sequence $(z_n)_{n=1}^{\infty}$ converges iff the real sequences formed by taking the real and imaginary parts of the original complex sequence both converge.

let $z_n=x_n+iy_n$
then $\sum_{n=1}^{\infty} z_n=\sum_{n=1}^{\infty} x_n+ i\sum_{n=1}^{\infty} y_n$

suppose that $\sum_{n=1}^{\infty} x_n$ and $\sum_{n=1}^{\infty} y_n$ converge

let $\epsilon>0$

$\sum_{n=1}^{\infty} x_n$ converges then $\exists N_1$ integer such that for every $n>=N_1$ we have $|\sum_{n=1}^{\infty} x_n-S| < \epsilon/2$

$\sum_{n=1}^{\infty} y_n$ converges then $\exists N_2$ integer such that for every $n>=N_2$ we have $|\sum_{n=1}^{\infty} y_n-L| < \epsilon/2$

Thus for $n>=max(N_1,N_2)$ we have

$|\sum_{n=1}^{\infty} z_n-(S+i L|<=|\sum_{n=1}^{\infty} x_n-S|+|\sum_{n=1}^{\infty} y_n-L|<\epsilon/2+\epsilon/2=\epsilon$

therefore $\sum_{n=1}^{\infty} z_n$ converges

Suppose now that $\sum_{n=1}^{\infty} z_n$ converges to $\gamma=\alpha +i\beta$

$|\sum_{n=1}^{\infty} x_n-\alpha|<=|\sum_{n=1}^{\infty} x_n-\alpha + i\sum_{n=1}^{\infty} y_n-\beta|=|\sum_{n=1}^{\infty} z_n-\gamma|$ thus $\sum_{n=1}^{\infty} x_n$ converges

$|\sum_{n=1}^{\infty} y_n-\beta|<=|\sum_{n=1}^{\infty} x_n-\alpha + i\sum_{n=1}^{\infty} y_n-\beta|=|\sum_{n=1}^{\infty} z_n-\gamma|$ thus $\sum_{n=1}^{\infty} y_n$ converges

@felyahia The question is about sequences - not series. If you work in that context, I think you'll avoid some crucial mistakes. For example, in lines 5 and 6, how can you have $n>N_{1\text{ or }2}$ if $n$ is an index of summation?

I think this reasoning works for series and for sequences as well, and you're right for n is an index of summation makes things messy. Can I rectify this mistake by using
$\sum_{k=1}^{n} $ instead of $\sum_{n=1}^{\infty} $ and make n go to infinity?