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In this solution, first it states that

If we write $f(z) = \frac{c}{z_0-z}+\sum_{n=0}^\infty b_nz^n$ then $\lim_{n\to\infty}b_n =0$.

I think I'm missing something very fundamental but why $\lim_{n\to\infty}b_n =0$? I mean if we think a geometric series $\sum_{n=0}^\infty z^n$, then this series converges on $|z|<1$. But its coefficient is $1$.

Also the limit

$$\lim_{n\to\infty}\frac{b_n+\frac{c}{z_0^{n+1}}}{b_{n+1}+\frac{c}{z_0^{n+2}}} = \lim_{n\to\infty}\frac{\frac{c}{z_0^{n+1}}}{\frac{c}{z_0^{n+2}}}$$

Could you explain this to me?

one potato two potato
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    The assumption is that $f$, except for the pole, is analytic on an open set containing the closed unit disc. So, removing the singular part at $z_0$ you get a series at $z=0$ with a radius of convergence strictly larger than $1$. Anyway, that answer is not very good. I would advice you not to follow it. – plop May 27 '21 at 01:35
  • If you look at the comments of that answer, some people have pointed out that (a) the assumption that $f$ can be written in that form with $b_n \to 0$ is not actually necessary, and (b) the limit at the end of your question needs to be justified more rigorously. I think TCL's answer attempts to address these gaps more rigorously. – angryavian May 27 '21 at 01:35
  • @angryavian For (a), in MSE, this question is frequently asked but many the answers assumed $\lim_{n\to\infty}b_n =0$ so I thought as an obvious fact lol – one potato two potato May 27 '21 at 01:38
  • @andreas It is a fact that follows from the radius of convergence being larger than $1$. You have that $\limsup_n|b_n|^{1/n}=r<1$. Therefore, for almost all $n$ we have that $|b_n|^{1/n}\leq (r+1)/2<1$. It follows that $0\leq|b_n|\leq ((r+1)/2)^n\to0$. Hence, $b_n\to0$. – plop May 27 '21 at 01:42
  • @plop How do you know $\limsup_n|b_n|^{1/n} =r<1$? As far as I understood, the expression $f(z) =\frac{c}{z_0-z}+\sum_{n=0}^\infty b_nz^n$ follows from the Laurent series expansion on a deleted nbd at $z_0$ contained in $B_R(0)$ where $R>1$. – one potato two potato May 27 '21 at 01:47
  • That is not a Laurent expansion around $z_0$. The justification of it is that, under the extra assumption that the pole is simple and that the residue is $-c$, then $f(z)-\frac{c}{z_0-z}$ is analytic in a neighborhood of the closed unit disc. They are expanding that difference around $z=0$. Otherwise you should be seeing a series of powers of $z-z_0$ instead of powers of $z$. – plop May 27 '21 at 01:51
  • @plop Is the fact that $f(z)-\frac{c}{z_0-z}$ is analytic in a neighborhood of the closed unit disc an obvious fact? How do you know that? – one potato two potato May 27 '21 at 01:55
  • Even if the radius of convergence is exactly $1$, as long as it converges at one point on the unit circle then $b_n \to 0$. – Erick Wong May 27 '21 at 02:04
  • @andreas It is a sum of analytic functions at all points where $f$ was analytic. At the point $z_0$ we are assuming that $(z-z_0)f(z)\to c$. Therefore $(z-z_0)f(z)-c$ has a removable singularity at $z=z_0$, which is actually a zero. Therefore, we can divide by $z-z_0$ and still be analytic. – plop May 27 '21 at 02:09
  • @plop So far I understood $(z-z_0)f(z)-c$ has a removable singularity at $z = z_0$. But why does it imply divide by $z-z_0$ and still analyic? – one potato two potato May 27 '21 at 04:52

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