TITLE: Let $a_n$ be a decreasing sequence. Prove that the power series $\sum a_n x^n$ has no roots in $A=\{z\in C:|z|<1\}$
QUESTION [5 upvotes]: Let $a_n$ be a decreasing, positive sequence, in the real space. Prove that the power series $\large p(z)=\sum\limits_{k=0}^{n} a_k z^k$ has no roots in $A=\{z\in C:|z|<1\}$. What I did so far

$zp(z) =\sum\limits_{k=1}^{n+1} a_{k} z^k \Rightarrow |zp(z)| \leq \sum\limits_{k=1}^{n+1} a_{k} |z|^k \leq\sum\limits_{k=0}^{n} a_k |z|^k = |p(z)|=|-p(z)|$ (since $a_n$ non decreasing and $|z|<1$)
using Rouche's principle $(z-1)p(z)$ has the same amount of roots with $-p(z)$ or $p(z)$. 

Am I going to the right direction ?

REPLY [6 votes]: The trick is a summation by parts. From
$$(1-z)p(z) = \sum_{k = 0}^{n} a_k z^k - \sum_{k = 1}^{n+1} a_{k-1} z^k = a_0 - \sum_{k = 1}^{n} (a_{k-1} - a_k)z^k - a_n z^{n+1}$$
we deduce that for $\lvert z\rvert < 1$ we have
\begin{align}
\lvert (1-z)p(z)\rvert &\geqslant a_0 - \sum_{k = 1}^{n} (a_{k-1} - a_k)\lvert z\rvert^k - a_n\lvert z\rvert^{n+1} \\
&> a_0 - \sum_{k = 1}^{n} (a_{k-1} - a_k) - a_n \\
&= 0\,.
\end{align}