# The convergence or divergence of the power series at a point does not determine whether the function can or cannot be continued beyond that point

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

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.

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 $$\lim_{z\to1}g''(z)=g''(1)$$.
• 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) Commented Dec 31, 2020 at 16:28
• 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} Commented Dec 31, 2020 at 17:15