Dgdrxrt

I'm looking to the prove of the following statement:

If for the complex power series $sum^infty_i=1 a_i z^i$ it holds that
$$ limsup_n to infty sqrt[n]<1, $$
then $sum^infty_i=1a_nz^n$ converges to the holomorphic function $f$ inside the disc $D =D(0,|z_0|)$ and the only singularity is $z_0$.

In the context of this question by singularity I mean there is no such holomorphic function $varphi$ defined in some neighborhood $U$ of $z_0$ such that $f_U cap D = varphi_U cap D$.

It is easy to see that in case for any $c in mathbbR_++$
$$
left|left n in mathbbN : leftright| = infty,
$$
the limit superior will be not less then $1$:
$$ lim_ntoinftysup left| fraca_na_n+1 -z_0 right|^1/n ge
lim_n to infty sqrt[n]c = 1. $$
This is a contradiction, hence there must be a convergence
$$
lim_n to infty fraca_na_n+1 = z_0,
$$
implying that the radius of convergence for $sum^infty_n=0a_nz^n$ is indeed $|z_0|$.

However, I don't know how to prove that $z_0$ is a singularity and any other point of $partial D$ is not.

Help me to prove that $z_0$ is a singular point and the other points in $partial D$ are not.

p. s.

This question seems to be a converse to About the limit of the coefficient ratio for a power series over complex numbers

I think I got it:

The series define holomorphic function $f(z) = sum^infty_n=0a_n z^n$ inside the disc $D =D(0,|z_0|)$. Consider the function $g(z) = (z - z_0)f(z)$. It can be expanded at the $0$ as $g(z) = -z_0a_0 + sum^infty_n=1(a_n-1 - z_0a_n)z^n$. Then it is possible to estimate a limit:
$$
limsup_n to inftysqrt[n]a_n-1 - z_0a_n =
limsup_ntoinfty sqrt[n]sqrt[n]left <
|z_0|^-1
$$
as it is known from before that $lim_ntoinftysup sqrt[n] = |z_0|^-1$. Hence, $g(z)$ is holomorphic inside the disc with the radius extending over the one of $D$. This is enough to claim that the series $sum^infty_n=1a_nz_n$ has only a 1st order pole in $z_0$ and regular point everywhere else on $partial D$, because function $varphi(z) = fracg(z)z - z_0$ will work as a sufficient analytic extension.