$f(z)=\frac{1}{1-z}=\sum\limits^{\infty}_{n=0}z^n\text{ for }\mid z\mid<1$.
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.
Best Answer
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 $\lim_{z\to1}g''(z)=g''(1)$.