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

(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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## closed as too broad by Did, Nosrati, Brahadeesh, Cesareo, Jaap ScherphuisAug 8 '18 at 15:59

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

by root or ratio test.

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 $$R \le 1$$.

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

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