Consider a holomorphic function $f(z)$ defined as power series $$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n},$$ with radius of convergence $R=1$.
I want to show that
If $a_{n}\geq 0$ for all $n$, then $f(z)$ cannot be analytically continued to $1\in \partial\mathbb{D}$.
I don't quite have an idea about where to start. As the exercise supposes $a_{n}\geq 0$ for all $n$, it means that the exercise also implicitly assumes that $a_{n}\in\mathbb{R}$ for all $n$.
From the radius of convergence, I know that $$\limsup_{n\rightarrow\infty}a_{n}^{\frac{1}{n}}=1\ \ \text{and}\ \ \lim_{n\rightarrow\infty}\dfrac{a_{n}}{a_{n+1}}=1.$$ (I get rid of the norm because $a_{n}$ is real and non-negative.)
Suppose $f(z)$ can be analytically continued to $1\in\partial\mathbb{D}$, then it means that there exists a $r=r(1)>0$ such that there exists a holomorphic function $g(z)$ on $B(1,r)$ such that $g(z)=f(z)$ on $B(1,r)\cap\mathbb{D}$.
But then I don't know how to proceed..
I tried to argue that if I define $h(z):=f(z)$ for $z\in\mathbb{D}$ and $h(z):=g(z)$ for $z\in B(1,r)$, then $h(z)$ is holomorphic in the whole $\mathbb{D}\cup B(1,r)$, so it has a power expansion $$h(z)=\sum_{n=0}^{\infty}b_{n}z^{n},$$ and in particular, at $z=1$, we must have $$\sum_{n=0}^{\infty}b_{n}<\infty.$$
But it seems no way to get back to $f(z)$. What should I do? Thank you!
Assuming $f$ is analytic at $1$, since everything is increasing $$f^{(k)}(1)=\lim_{z\to 1^-}f^{(k)}(z)= \lim_{z\to 1^-} \sum_{n=k}^\infty a_n \frac{n!}{(n-k)!} z^{n-k}=\sum_{n=k}^\infty a_n \frac{n!}{(n-k)!}$$ Next we are told that for any $s>0$ $$\sum_{n=0}^\infty a_n (1+s)^n= \infty$$ Everything is non-negative so we can change the order of summation as we want obtaining $$\infty=\sum_{n=0}^\infty a_n \sum_{k=0}^n s^k\frac{ n!}{k!(n-k)!} = \sum_{k=0}^\infty \frac{s^k}{k!} \sum_{n=k}^\infty a_n\frac{n! }{(n-k)!}=\sum_{k=0}^\infty \frac{s^k}{k!} f^{(k)}(1)$$ contradicting that $f$ is analytic at $1$.