A sequence $\{z_n\}$ converges if $\lim\limits_{n\to\infty}z_n=L$, where $L$ is a complex number. In other words, $\text{Re}(z_n)$ converges to $\text{Re}(L)$, and $\text{Im}(z_n)$ converges to $\text{Im}(L)$.
Section 19.1: 3, 4, 7, 9, 11, 14, 18, 19, 22, 23, 27, 29.
A sequence $\{z_n\}$ converges if $\lim\limits_{n\to\infty}z_n=L$, where $L$ is a complex number. In other words, $\text{Re}(z_n)$ converges to $\text{Re}(L)$, and $\text{Im}(z_n)$ converges to $\text{Im}(L)$.
For instance, the sequence ${1+i^n}$ does not converge because its terms do not approach a fixed complex number as $n$ goes to infinity
Is is convergent if the sequence of partial sums, $\{S_n\}$converges:
$$ \sum_{k=1}^{\infty}z_k = z_1 + z_2 + z_3 + ... + z_n + ... $$
If $S_n \to L$ as $n \to \infty$, then we say the sum of the series is L.
Defined as:
$$ \sum_{k=1}^{\infty}az^{k-1}=a+az+az^2+az^3+... $$
When $|z| < 1$, the series converges to the following, and diverges for $|z| ≥1$
$$ S_\infty = \frac{a}{1-z} $$
Ratio Test:
$$ \lim\limits_{n\to\infty}\left|\frac{z_{n+1}}{z_n}\right| = L $$
Root Test:
$$ \lim\limits_{n\to\infty}\sqrt[n]{|z_n|}=L
$$
A power series centred at $z_0$ is of the form below, where $a_k$ coefficient is a complex constant.
$$ \sum_{k=1}^{\infty}a_k(z-z_0)^k $$
Circle of convergence is a circle of radius $R$ centered at $z_0: |z-z_0| = R$
$$ (i) \quad \lim_{{n \to \infty}} \left| \frac{{a_{n+1}}}{{a_n}} \right| = L \neq 0, \text{ the radius of convergence is } R = \frac{1}{L};
$$
$$ (ii)\quad \lim_{{n \to \infty}} \left| \frac{{a_{n+1}}}{{a_n}} \right| = 0, \text{ the radius of convergence is } R = \infty; $$
$$ (iii) \quad \lim_{{n \to \infty}} \left| \frac{{a_{n+1}}}{{a_n}} \right| = \infty, \text{ the radius of convergence is } R = 0. $$
