Let $\sum_{n=0}^\infty a_n z^n$ be a formal complex power series, with $a_n$ strictly decreasing to 1 as $n\to \infty$. It is easy to see that the radius of convergence is 1, so that this power series represents a function $f$ holomorphic on the disk. The question is to
Show that f has no zeroes on the disk.
I've been looking at this a while and feel and tried a few different things.
$\cdot$ One easy observation is the similarity of this series with $\frac{1}{1-z}$, whose coefficients do not strictly decrease but which otherwise is an example of the thing to be proven.
$\cdot$ Rouché's theorem doesn't seem to have an immediate application—I've tried comparing $f$ to $\frac{1}{1-z}$ or to its partial sums by Rouché.
$\cdot$ If the partial sums had no zeroes, then neither would $f$ by Hurwitz's theorem
$\cdot$ If we could show the function had positive real part we'd obviously be done, which seems highly plausible—the condition gives that $f(-1) = \sum a_n \cos\pi n > 0$, but I don't know how to extend this argument to arguments other than $\pi$. Besides, if this technique were to work, it seems it would rely most on algebraic manipulations of power series, and I would vastly prefer a classical complex analytic approach (argument principle, Rouché's theorem, etc)
Please advise!
Let us consider $f(z) =(1-z)\sum\limits_{n=0}^\infty a_n z^n =a_0+\sum\limits_{n=1}^\infty (a_n-a_{n-1})z^n$. Suppose $f(re^{i\theta})=0$ for $r\le 1$. Then it follows $$ a_0 =\sum_{n=1}^\infty (a_{n-1}-a_n)r^ne^{in\theta}, $$ hence $$ a_0=\left|\sum_{n=1}^\infty (a_{n-1}-a_n)r^ne^{in\theta}\right|\le \sum_{n=1}^\infty (a_{n-1}-a_n)r^n\le\sum_{n=1}^\infty(a_{n-1}-a_n)=a_0-1, $$ which leads to a contradiction. Since $f$ has no zeros on the unit disk, neither does $\frac{f(z)}{1-z}=\sum\limits_{n=0}^\infty a_n z^n$.