About the Radius of Convergence of $ \sum_{n\ge 0}a_n z^n $ 
Fix $ \delta>0 $ and let
  $$ \Omega=\{ z\in\mathbb C:|z|<1 \}\cup\{ z\in\mathbb C:|z-1|<2\delta \} .$$
  Assume that $ f(z) $ is a holomorphic function on $ \Omega $ whihc has a Taylor series expansion $ \sum_{n\ge 0}a_nz^n $ at $ z=0 $ such that $ a_n $ is a non-negative real number for all $ n\ge 0 $.
(A) Prove that the derivatives $ f^{(k)}(1) $ are real for all $ k\ge 0 $ and, moreover, $$ f^{(k)}(1)\ge\frac{m!}{(m-k)!}a_m $$ for all $ 0\le k\le m $.
(B) Prove that $ \sum_{n\ge 0}a_n z^n $ has radius of convergence strictly greater than $ 1 $.


My attempt:
(A) Since $ f(z)=\sum_{n\ge 0}a_nz^n $ when $ |z|<1 $, we have
\begin{align}
f(z)&=\sum_{n\ge 0}a_n[(z-1)+1]^n\\
&=\sum_{n\ge 0}a_n\sum_{m=0}^n\binom{n}{m}(z-1)^m\\
&=\sum_{k\ge 0}\left[\sum_{n\ge k}a_n\binom{n}{k}\right](z-1)^k\\
&=\sum_{k\ge 0}\frac{f^{(k)}(1)}{k!}(z-1)^k
\end{align}
for $$ z\in\{z:|z|<1\}\cap\{ z:|z-1|<2\delta \} .$$
\begin{align}
&\implies \frac{f^{(k)}(1)}{k!}=\sum_{n\ge k}a_n\binom{n}{k}\\
&\implies\frac{f^{(k)}(1)}{k!}=\sum_{n\ge k}a_n\frac{n!}{k!(n-k)!}\\
&\implies f^{(k)}(1)=\sum_{n\ge k}a_n\frac{n!}{(n-k)!}
\end{align}
Hence $ f^{(k)}(1) $ are real for all $ k\ge 0 $ and since $ a_n\ge 0 $, we have
$$ f^{(k)}(1)\ge\frac{m!}{(m-k)!}a_m\quad\text{for all}\ 0\le k\le m .$$ So we have proved (A).
(B) Since there must exist at least one singular point on the boundary of the disk of convergence, it suffices to prove that there exists a singular point on $ \{ z\in\mathbb C:|z|=1 \} $. Then I am stuck...... Any hint?
 A: Note that if $|z| < 1+\delta$, \begin{align}\sum{|a_nz^n|} &\le \sum{a_n(1+\delta)^n}\\
&=\sum_{n \ge 0}\left(a_n\sum_{k=0}^{n}\binom{n}{k}{\delta}^k\right)\\
&=\sum_{k\ge 0}\left({\delta}^k\sum_{n\ge k}a_n\binom{n}{k}\right)\\
&=\sum_{k=0}^\infty\frac{f^{(k)}(1)}{k!}\,\delta^k < \infty,\end{align} hence $f(z)$ has a Taylor series defined in the (open) disc with radius $1+\delta$
The switching of the double sum is allowed by the non-negativity of all terms as $a_{mn}\ge 0$ implies $$\sum_{m}(\sum_{n}a_{mn})=\sum_{n}(\sum_{m}a_{mn})=\sup_{n \in I, m \in J}\sum{a_{mn}}$$ with supremum taken on all finite sets $I,J$ of natural numbers, the double sums being finite and equal or both infinite
Note that this result is also known as the power series version of Landau's Theorem (Landau's Theorem being much better known for Dirichlet series where it is slightly more difficult to prove than here), stating that if a power series with radius of convergence precisely $r>0$ has non-negative coefficients (for all $n$ high enough), than it must have a singularity at $z=r$. In particular here $r=1$ cannot be the radius convergence of $f$ as $1$ is not a singular point by hypothesis.
A: Maybe useful idea, too long for a comment: as $f$ is holomorphic in $\{z\in\Bbb C:|z - 1| < 2\delta\}$ the Taylor series of $f$ centered at $1$
$$\sum_{n=0}^\infty\frac{f^{(n)}(1)}{n!}(z - 1)^n$$
is convergent at $z = 1 + \delta$, i.e.,
$$\sum_{n=0}^\infty\frac{f^{(n)}(1)}{n!}\,\delta^n$$
is convergent. The convergence of this series plus (A) maybe give a useful bound for $a_n$.
