Find radius of of $\sum k^n z^k$, $\sum \frac{z^k}{k^k}$, $\sum \cos(k)z^k$, $\sum 4^k (z-2)^k$

Question

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 7.33(b)(e)(f)(g)

(b) What's the role of $n \in \mathbb Z$? I guess it's in the part where $\lim_{k} |\frac{k+1}{k}|^n = |\lim_{k}\frac{k+1}{k}|^n$ like $x^n$ is continuous if $n \in \mathbb Z$?

(e) $R = \infty$? WA just says converges by ratio test (I used root).

(f) $R=1$? I use Exer 7.27 (*) to say that since $|\cos(k)| \le 1 \ \forall k \ge 0$ and $\lim_k \cos(k) \ne 0$, we have resp, $R \ge 1$ and $R \le 1$.

(g) $R=\frac 1 4$? WA initially says that series converges for the equivalent of $|z-2| < \frac 1 4$ but then later says that the series diverges by the 'geometric series test'.

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Answer 1

(b) If $n \in \mathbb Z$, then $z^n$ is continuous. Thus, $\lim_{k} |\frac{k+1}{k}|^n = |\lim_{k}\frac{k+1}{k}|^n$.

(e) $R = \infty$

by root or ratio test.

(f) $R=1$

By Exer 7.27 (*), since $|\cos(k)| \le 1 \ \forall k \ge 0$ and $\lim_k \cos(k) \ne 0$, we have resp, $R \ge 1$ and $R \le 1$.

(g) $R=\frac 1 4$.

Power series usually diverge. It's just that that converge for some subset of $\mathbb R$ or $\mathbb C$.

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