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[1]The series is $$\sum_{n=1}^\infty \frac{2^n(1-i)^\left(2n\right)}{(1+i)^\left(n+1\right)}*(z+i)^n$$.

If i use the ratio rule it becomes$$\sum_{n=1}^\infty \frac{2^\left(n+1\right)(1-i)^\left(2n + 1\right)}{(1+i)^\left(n + 2\right)}*(z+i)^\left(n + 1\right)$$, I think but I'm not sure how to go from here

HokieFan7
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1 Answers1

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In order to apply the ratio test, you are supposed to compute the limit$$\lim_{n\to\infty}\frac{\left|\frac{2^{n+1}(1-i)^{2n+2}}{(1+i)^{n+2}}(z+i)^{n+1}\right|}{\left|\frac{2^n(1-i)^{2n}}{(1+i)^{n+1}}(z+i)^n\right|}.\tag1$$But\begin{align}(1)&=\lim_{n\to\infty}\frac{2|1-i|^2}{|1+i|}|z+i|\\&=\frac{2\times2}{\sqrt2}|z+i|\\&=2\sqrt2|z+i|,\end{align}and therefore the radius of convergence is $\frac1{2\sqrt2}$.

José Carlos Santos
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  • Thanks again for your answer. Can you explain what my teacher is doing, because I can see what you did but looking over the notes I have, i get lost again. I edited my original question and included a link to an example from my notes. In his example, he didn't include the (z - i) term and I'm not really sure why – HokieFan7 Apr 27 '20 at 21:06
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    That's because your teacher taught you that the radius of convergence of the power series $\sum_{n=0}^\infty a_nz^n$ is $\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|$, when this limit exits. – José Carlos Santos Apr 27 '20 at 21:12
  • Oh right I forgot this one starts at 1 not 0, thanks so much again – HokieFan7 Apr 27 '20 at 21:18
  • I just realized I accidentally used i as my index instead of n and n actually starts at 1 and goes to inf, not 0. Does that end up changing the answer? – HokieFan7 Apr 27 '20 at 21:26
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    Where $n$ starts doesn't affect the radius of convergence. – José Carlos Santos Apr 27 '20 at 21:31
  • Just one final question, in the (1-i) term in the numerator, why is it to the 2n+2 power, when you add 1 to the original series doesn't it become 2n+1? – HokieFan7 Apr 27 '20 at 21:40
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    Because $(1-i)^{2(n+1)}=(1-i)^{2n+2}$. – José Carlos Santos Apr 27 '20 at 21:44
  • i see that but in the original problem the exponent for (1-i) is 2n. Did you combine that with the 2^(n+1) next to it and that's why it becomes 2n+2? – HokieFan7 Apr 27 '20 at 21:48
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    No. It's because, as I wrote in my previous comment, $(1-i)^{2(n+1)}=(1-i)^{2n+2}$. So, if $a_n=\frac{2^n(z-i)^{2n}}{(1+i)^{n+1}}(z-i)^n$, then $a_{n+1}=\frac{2^n(z-i)^{2n+2}}{(1+i)^{n+2}}(z-i)^{n+1}$. – José Carlos Santos Apr 27 '20 at 22:03