Use the Weierstrass M-test to show that each of the following series converges uniformly on the given domain:
a) $\displaystyle \sum \frac{z^k}{k^2} \text{ on } \overline{D}_1(0)$
b) $\displaystyle \sum \frac{1}{z^k} \text{ on } \overline{D}_2(0)^c$
c) $\displaystyle \sum \frac{z^k}{z^k+1} \text{ on } \overline{D}_r(0) \text{ where } 0\leq r<1$
For a.) We know that $f_k(z)=\frac{z^k}{k^2}$ and that $\frac{1}{k^2}\leq1$ for $k\geq1.$ So we know that $\frac{z^k}{k^2}\leq z^k=M_n$ in our given domain. So, by the Weierstrass M-test, if we can conclude that $z^k$ converges, then $\frac{z^k}{k^2}$ must also converge. Luckily, we know the $\sum z^k$ converges to $\frac{1}{1-z},$ so $\sum\frac{z^k}{k^2}$ converges under our given domain.
@opernie Pretty good - just a couple little issues:
First, how do we know $z^k/k^2 \leq 1/k^2$? (I agree, but it is worth referring to the domain.)
More importantly, why the reference to the geometric series?
For c, I used a conjugate, which worked because for $r<1$, I started with:
$$\frac{z^k-1}{z^k-1}\cdot \frac{z^k}{z^k+1}=z^k\cdot \frac{z^k-1}{z^{2k}-1}$$
For $r<1$, we know that:
$$z^k\cdot \frac{z^k-1}{z^{2k}-1}<z^k$$
If we let $\textbf{M}_k=z^k$, because $\sum z^k$ converges, we know $\sum \frac{z^k}{z^k+1}$ converges.