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 ?


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}


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