# 2 Exercises on Weierstrass M-test for power series

Exer 7.29

(c) $M_k = r^k$ because $$\frac{|z|^k}{r^k} \le 1 \le |z^k + 1|$$

Elaboration: I believe that $|\frac{z^k}{z^k+1}| \le r^k$.

Pf: For $|z| \le r,$ we have $$|z|^k \le r^k \implies \frac{|z|^k}{r^k} \le 1. \tag{1}$$

But for $|z| \in [0,\infty),$ $$|z^k+1| \ge ||z^k|-|1|| = ||z|^k-1| \ge |0-1| = 1 \implies 1 \le |z^k+1| \tag{2}$$

Thus,

$$(1) \wedge (2) \implies \frac{|z|^k}{r^k} \le 1 \le |z^k+1| \implies \frac{|z|^k}{|z^k+1|} \le r^k$$

$$\therefore, |\frac{z^k}{z^k+1}| = \frac{|z^k|}{|z^k+1|} = \frac{|z|^k}{|z^k+1|} \le r^k \ \text{QED}$$

Exer 7.30

Weierstrass M-Test: $|\frac{z}{w}|^k \le \frac{|z|^k}{r^k} = |\frac{z}{r}|^k =: M_k$

In exercise 7.29 (c) $$\left\lvert \frac{z^k}{z^k+1} \right\rvert \leq r^k$$ doesn't hold for all of the $z \in \overline{D}[0,r]$. According to the maximum modulus principle, the maximum occurs on the boundary where $z=r \exp{i \theta}$. The functions to be maximized then become $$\frac{r^k}{|r^k \exp{i k \theta}+1|},$$ and the maxima occur when $\exp{i k \theta}=-1$. Hence $M_k$ really should be $$M_k:=\frac{r^k}{1-r^{k}} .$$

Exericse 7.30 looks ok to me.

• Thanks user1337, but what exactly is wrong with my argument please? – BCLC Aug 4 '18 at 2:26
• @BCLC Your choice of $M_k$. – user1337 Aug 4 '18 at 17:22
• lol I mean why? I believe it's an upper bound whose series is convergent – BCLC Aug 4 '18 at 20:11
• @BCLC It is not an upper bound. Consider $z=-r$ when $k$ is odd, for instance. – user1337 Aug 4 '18 at 21:02
• @BCLC $||z|^k-1|\ge|0-1|$ is not necessarily true, consider $z=0.1$ – user341124 Aug 5 '18 at 14:08

$$\frac{r^k}{1-r^{k}} \le \frac{r^k}{1-r}$$
$$|\frac{z^k}{1+z^k}| \le \frac{r^k}{1-r}$$
Note: Upper bound is a my proposed (albeit incorrect) upper bound multiplied by $\frac{1}{1-r}$. Note that $\frac{1}{1-r}$ does not depend on $k$.