Proposition

Use the Weierstrass M-test to show that each of the following series converges uniformly on the given domain.

$\sum_{k \geq 1} \frac{z^k}{k^2}$ on $\overline{D}[0, 1]$.
$\sum_{k \geq 0} \frac{1}{z^k}$ on $\{ z \in \mathbb{C} : \abs{z} \geq 2 \}$.
$\sum_{k \geq 0} \frac{z^k}{z^k + 1}$ on $\overline{D}[0, r]$ where $0 \leq r < 1$.

Solution

1

Let $M_k = 1/k^2$. Then $\abs{\frac{z^k}{k^2}} \leq M_k$ for all $k$ and $z \in \overline{D}[0, 1]$, and $\sum M_k$ converges.

2

Let $M_k = 1/2^k$. Ten $\abs{\frac{1}{z^k}} \leq M_k$ for all $k$ and $z$ in the specified domain.

3

Let $M_k = \frac{r^k}{1 - r}$.

We have $1 = \abs{1 + z^k - z^k} \leq \abs{1 + z^k} + \abs{z^k}$. Then $1 - \abs{z^k} \leq \abs{1 + z^k}$. Since $\abs{z^k} = \abs{z}^k \leq r$, $1 - r \leq \abs{1 + z^k}$.

Therefore, $\abs{\frac{z^k}{z^k + 1}} \leq \frac{r^k}{1 - r} = M_k$.