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?

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 $a_n$.
