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I am trying to find the radius of convergence and trying to figure out the behaviour on the frontier of the disk of convergence of the following power series:

a) $\sum_{n=1}^{\infty} \dfrac{n!}{(2-i)n^2}z^n$

b) $\sum_{n=1}^{\infty} \dfrac{1}{1+(1+i)^n}z^n$

I know that the radius of convergence of a power series is $R$ where $\dfrac{1}{R}=\overline{\sup \lim} \sqrt[n]{|a_n|}$

So, in a), we have $\overline{\sup \lim} \sqrt[n]{|a_n|}=\dfrac{\sqrt[n]{n!}}{\sqrt[n]{\sqrt{5}n^2}}$, I have no idea how to calculate this limit. And in b) I have the same problem, how do I calculate $\lim_{n \to \infty} \dfrac{1}{\sqrt[n]{|1+(1+i)^n|}}$

I would really appreciate help calculating these limits. Thanks in advance.

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One would rather use the ratio test.

For a), one obtains, as $n \to \infty$, $$ \left|\dfrac{(n+1)!}{(2-i)(n+1)^2}\times \dfrac{(2-i)n^2}{n!}\right|=\frac{n^2}{(n+1)} \to \infty $$ thus $R=0$.

For b), one obtains, as $n \to \infty$, $$ \left|\dfrac{1}{1+(1+i)^{(n+1)}}\times \dfrac{1+(1+i)^n}{1}\right|=\left|\dfrac{1}{1+i}\right| \times \left|\dfrac{1+(1+i)^{-n}}{1+(1+i)^{-(n+1)}}\right|\to \frac1{\sqrt{2}} $$ thus $R=\sqrt{2}$. On the frontier of the disk of convergence, one may write $z=\sqrt{2}e^{i\theta}$, $\theta \in [0,2\pi]$ then one may observe the behaviour of the general term, $$ \dfrac{\left(\sqrt{2}\right)^n}{1+(1+i)^n}\,e^{in\theta} $$ giving, as $n \to \infty$: $$ \left|\dfrac{\left(\sqrt{2}\right)^n}{1+(1+i)^n}\,e^{in\theta}\right|=\left|\dfrac{\sqrt{2}}{1+i}\right|^n \times \left|\dfrac1{1+(1+i)^{-n}}\right|\to 1 \neq0, $$ the series is divergent everywhere on the frontier of the disk.

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  • $\begingroup$ OK, but OP also wants to know the behavior on the circle of radius $R$. $\endgroup$ – Gerry Myerson May 6 '16 at 7:20
  • $\begingroup$ Edited. Thanks. $\endgroup$ – Olivier Oloa May 6 '16 at 9:58
  • $\begingroup$ Sorry to ask you after accepting the answer, but now that I read the solution in detail I realize I don't understand why $\dfrac{1+(1+i)^n}{1+(1+i)^{n+1}}=\dfrac{1}{1+i}(\dfrac{1+(1+i)^{-n}}{1+(1+i)^{-(n+1)}})$ and why the absolute value of the last expression tends to $\dfrac{1}{\sqrt{2}}$, could you explain these steps in further detail? $\endgroup$ – user156441 May 6 '16 at 23:40
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    $\begingroup$ @user156441 Sure. Observe that $1+(1+i)^n=(1+i)^n\left(\frac1{(1+i)^n}+1\right)$ and this equals $(1+i)^n\left((1+i)^{-n}+1\right)=(1+i)^n\left(1+(1+i)^{-n}\right)$. Similarly, $1+(1+i)^{n+1}=(1+i)^{n+1}\left(1+(1+i)^{-(n+1)}\right)$. Thus, since $\frac{(1+i)^n}{(1+i)^{n+1}}=\frac{1}{(1+i)}$ then $\frac{1+(1+i)^n}{1+(1+i)^{n+1}}=\frac1{1+i}(\frac{1+(1+i)^{-n}}{1+(1+i)^{-‌​(n+1)}})$. Observe that $|1+i|=\sqrt{2}$ and that $(1+i)^{-n} \to 0$. That's why the mentioned expression tends to $1/\sqrt{2}$. Hoping these details help! $\endgroup$ – Olivier Oloa May 7 '16 at 7:37

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