Let $(a_n)_{n \geq 0}$ be a strictly decreasing sequence of positive real numbers , and let $z \in \mathbb C$ , $|z| < 1$.

Let $(a_n)_{n \geq 0}$ be a strictly decreasing sequence of positive real numbers , and let $z \in \mathbb C$ , $|z| < 1$. Prove that the sum $a_0 + a_1z + a_2z^2 + \cdots + a_nz^n +\cdots $ is never equal to zero.

Solution 1:

Simply note that $(1-z)\sum_{k=0}^\infty a_kz^k=a_0+\sum_{k=1}^\infty (a_k-a_{k-1})z^k$.

If $\sum_{k=0}^\infty a_kz^k=0$ for some $|z|<1$, $$a_0=\sum_{k=1}^\infty (a_{k-1}-a_k)z^k\implies$$ $$ \begin{align} a_0\leq \left| \sum_{k=1}^\infty (a_{k-1}-a_k)z^k\right| &\leq \sum_{k=1}^\infty (a_{k-1}-a_k)|z|^k \\ &< \sum_{k=1}^\infty (a_{k-1}-a_k)\\ &= a_0 - \lim_n a_n \end{align}$$

Hence $\lim a_n<0$ which is a contradiction.

This has some interesting consequences, like $\frac{1}{1-z}+e^z$ has no zeroes if $|z|<1$...

Solution 2:

Let us denote $f(x) = a_0 + a_1 x + a_2 x^2 + \cdot \cdot \cdot + a_n x^n + \cdot \cdot \cdot$. As $(a_n)$ is a decreasing sequence of positive real numbers, $f$ has a radius of convergence $\geqslant 1$.

We take $z$ such that $\mid z \mid <1$ (hence all the series that will be used are absolutely convergent). We assume $f(z) = 0$.

We have $(1-z)f(z) = 0$, so $a_0 = (a_0 - a_1)z + (a_1 - a_2)z^2 + (a_2 - a_3)z^3 + \cdot \cdot \cdot$

We use triangular inequality : $\mid a_0 \mid = \mid \sum \limits_{n=0}^{+\infty} (a_n - a_{n+1})z^{n+1} \mid \leqslant \sum \limits_{n=0}^{+\infty}\mid a_n - a_{n+1} \mid \cdot \mid z \mid ^{n+1}$.

Moreover, $a_0 > 0$ and $(a_n)$ is decreasing, so :

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$a_0 \leqslant \sum \limits_{n=0}^{+\infty} (a_n - a_{n+1})\mid z\mid ^{n+1}$ that is to say $a_0 \leqslant a_0 \mid z \mid - \sum \limits_{n=0}^{+\infty} a_{n+1} (\mid z \mid ^{n+1} - \mid z \mid ^{n+2})$.

$ $

Also $a_n > 0$ and $0 < \mid z \mid$ so this last sum is $>0$. We have finally $a_0 < a_0 \mid z \mid$ but $a_0 > 0$ and $\mid z \mid < 1$. Our assumption was wrong, so $\mid z \mid \geqslant 1$.