Dgdrxrt

Fix $ delta>0 $ and let
$$ Omega=<1 cupz-1 .$$
Assume that $ f(z) $ is a holomorphic function on $ Omega $ whihc has a Taylor series expansion $ sum_nge 0a_nz^n $ at $ z=0 $ such that $ a_n $ is a non-negative real number for all $ nge 0 $.

(A) Prove that the derivatives $ f^(k)(1) $ are real for all $ kge 0 $ and, moreover, $$ f^(k)(1)gefracm!(m-k)!a_m $$ for all $ 0le kle m $.

(B) Prove that $ sum_nge 0a_n z^n $ has radius of convergence strictly greater than $ 1 $.

My attempt:

(A) Since $ f(z)=sum_nge 0a_nz^n $ when $ |z|<1 $, we have

beginalign
f(z)&=sum_nge 0a_n[(z-1)+1]^n\
&=sum_nge 0a_nsum_m=0^nbinomnm(z-1)^m\
&=sum_kge 0left[sum_nge ka_nbinomnkright](z-1)^k\
&=sum_kge 0fracf^(k)(1)k!(z-1)^k
endalign
for $$ zinzcap z: .$$
beginalign
&implies fracf^(k)(1)k!=sum_nge ka_nbinomnk\
&impliesfracf^(k)(1)k!=sum_nge ka_nfracn!k!(n-k)!\
&implies f^(k)(1)=sum_nge ka_nfracn!(n-k)!
endalign

Hence $ f^(k)(1) $ are real for all $ kge 0 $ and since $ a_nge 0 $, we have
$$ f^(k)(1)gefracm!(m-k)!a_mquadtextfor all 0le kle m .$$ 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 $ zinmathbb C: $. Then I am stuck...... Any hint?

Note that if $|z| < 1+delta$, beginalignsuma_nz^n &le suma_n(1+delta)^n\
&=sum_n ge 0left(a_nsum_k=0^nbinomnkdelta^kright)\
&=sum_kge 0left(delta^ksum_nge ka_nbinomnkright)\
&=sum_k=0^inftyfracf^(k)(1)k!,delta^k < infty,endalign 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 $a_mnge 0$ implies $$sum_m(sum_na_mn)=sum_n(sum_ma_mn)=sup_n in I, m in Jsuma_mn$$ 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.

Maybe useful idea, too long for a comment: as $f$ is holomorphic in $zinBbb C:$ the Taylor series of $f$ centered at $1$
$$sum_n=0^inftyfracf^(n)(1)n!(z - 1)^n$$
is convergent at $z = 1 + delta$, i.e.,
$$sum_n=0^inftyfracf^(n)(1)n!,delta^n$$
is convergent. The convergence of this series plus (A) maybe give a useful bound for $a_n$.