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

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(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 Scherphuis Aug 8 '18 at 15:59

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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

(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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