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I apologize in advance for the simplicity of this question, I'm sure I'm missing something right in front of me.

For $z < r $, how can we show that the following limit evaluates to zero?

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We have $\alpha = |z/r| <1$, so $\alpha = 1/(1+\beta)$ where $\beta > 0$.

For all $n >2$, by the binomial theorem,

$$0 \leqslant n\alpha^{n-1} = \frac{n}{(1+ \beta)^{n-1}} < \frac{n}{\frac{1}{2}(n-1)(n-2)\beta^2} \rightarrow_{n \to \infty} 0 $$

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If $a<0$,$$\lim_{x\to\infty}xe^{a(x-1)}=\lim_{x\to\infty}\frac x{e^{-a(x-1)}}=0,$$by L'Hopital's rule. Now, apply this with $a=\log\left\lvert\frac zr\right\rvert<0$.

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  • $\begingroup$ I tried this approach first, if I'm not mistaken the second limit has infinity over zero :/ $\endgroup$ – Math Enthusiast Aug 15 at 14:08
  • $\begingroup$ Since $a<0$, it is actually $\frac\infty\infty$. And then you apply L'Hopital's Rule. $\endgroup$ – José Carlos Santos Aug 15 at 14:09
  • $\begingroup$ Oh, thank you very much! $\endgroup$ – Math Enthusiast Aug 15 at 20:41

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