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Let the serie $\sum_{k \geq 0} a_k (z-i)^k$ converge for $z = 4$ and diverge for $z=-8$. What is known of convergence and divergence of the following series?

(a) $\sum_{k \geq 0} a_k (1+i)^k$

(b) $\sum_{k \geq 0} a_k 9^k$

(c) $\sum_{k \geq 0} (-1)^k a_k 5^k$

A good criterion I could use here would be the radius of convergence (respectively, radius of "divergence").To do this, I thought to use the following theorem, but I'm not sure whether it is the right solution.

Theorem (Cauchy): Given a power serie with complexe coefficients $a_k$, $$\sum_{k \geq 0} a_k z^k,$$ let $$R= \frac{1}{\lim \sup_k |a_k|^{1/k}}$$ ($0 \leq R \leq \infty$). Then the serie converge absolutely in the ball $B(0,r)$, uniformly over the entire ball $\bar{B}(0,r)$, such that $r<R$, and it diverge if $|z|>R$

A part of the answer say we have convergence inside $B(i,\sqrt{17})$ and diverge outside $B(i,\sqrt{65})$. We don't have information between both ball.

Is anyone could explain to me how to obtain this radius of convergence?

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    $\begingroup$ You should rather using the (simpler looking facts) that if a series $\sum a_nx^n$ has radius of convergence $R$, then $\sum a_nx^n$ converges for every $x$ such that $|x|<R$ and $\sum a_nx^n$ diverges for every $x$ such that $|x|>R$, thus if $\sum a_nx^n$ converges then $|x|\leqslant R$ and if $\sum a_nx^n$ diverges then $|x|\geqslant R$. Thus, the hypothesis that $\sum a_n(4-i)^n$ converges and that $\sum a_n(-8-i)^n$ diverges tell you that $R$ is such that... hence... $\endgroup$
    – Did
    Jan 25 '16 at 12:36
  • $\begingroup$ I'm sorry, but I tried to solve with your hint, but I am not able to obtain the balls $B(i \sqrt{17})$ and $B(i \sqrt{65})$. Could you be a bit more precise? $\endgroup$ Jan 25 '16 at 13:11
  • $\begingroup$ Sure. Which information on $R$ did you get from my first comment? $\endgroup$
    – Did
    Jan 25 '16 at 13:14
  • $\begingroup$ For (a) $|1+i|=√2<√17$ which implies that $z=1+2i$ lies inside the circle of convergence. $\endgroup$ Jan 25 '16 at 13:15
  • $\begingroup$ @Did I am probably needed of a radius of convergence equal to $R-i$. The problem is I don't know how to obtain the particular R. $\endgroup$ Jan 25 '16 at 13:17
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The radius of convergence $r$ is at least $\sqrt {17}$ and at most $\sqrt {65}.$ We have convergence in (a) because $|1+i|=\sqrt 2<\sqrt {17}< r$ and divergence in (b) because $|9|=9>\sqrt {65}>r$. We do not have sufficient information to decide (c).

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