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$f(z)=\frac{1}{1-z}=\sum\limits^{\infty}_{n=0}z^n\text{ for }\mid z\mid<1$.

Although the power series diverges at every point on the unit circle, $f$ is analytic throughout the punctured plane $z\neq 1$.

How do I show it diverges at every point on unit circle ? How do I see the analyticity? Also it seems to contradict to the definition of "analyticity" because "analytic at a point" means we have a power series at that point with a positive radius of convergence, which is impossible (cause beyond $\mid z\mid=1$, the power series diverges)

$\sum\limits^{\infty}_{n=1}(z^n/n^2)$ converges at all points on the unit circle; however $g(z)$ cannot be continued analytically to a domain including $z=1$ since

$g''(z)=\sum\limits^{\infty}_{n=0}\frac{(n+1)z^n}{n+2}\to \infty\text{ > as }z\to 1^-$

I see the convergence on the unit circle. But how do we see the analyticity and why is second derivative relevant? I'd appreciate any insight.

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If $|z|=1$, then you don't have $\lim_{n\to\infty}z^n=0$ (since $(\forall n\in\Bbb N):|z^n|=1$) and therefore the series $\sum_{n=0}^\infty z^n$ diverges. And $f$ is analytic since it is the quotient of two analytic functions.

Concerning $g$, if you could expand it to analytically to a domain including $1$, the same would happen to $g''$. But then you would have $\lim_{z\to1}g''(z)=g''(1)$.

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  • $\begingroup$ Thanks for your response! Could you spare some time to address my doubts on the definition of analyticity? From my understanding, "analytic at a point p" means that there is a power series centered at a point p with a positive radius of convergence. But in the first example, we say, for example, $f$ is analytic at a point -1. Wouldn't it contradict to the definition because the power series centered at a point -1 cannot have a positive radius of convergence? (specifically, points to the left of -1 will diverge) $\endgroup$
    – able20
    Commented Dec 31, 2020 at 16:28
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    $\begingroup$ Note that we have\begin{align}f(z)&=\frac1{1-z}\\&=\frac1{2-(z+1)}\\&=\frac12\times\frac1{1-\frac{z+1}2}\\&=\frac12\sum_{n=0}^\infty\left(\frac{z+1}2\right)^n\text{ (when $|z+1|<2$)}\\&=\sum_{n=0}^\infty\frac{(z+1)^n}{2^{n+1}}.\end{align} $\endgroup$ Commented Dec 31, 2020 at 17:15
  • $\begingroup$ Thank you!! It does have a positive radius of convergence. I think I was confused about some facts about power series. $\endgroup$
    – able20
    Commented Dec 31, 2020 at 17:19

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