Proof that a necessary condition for characteristic roots to lie inside unit circle is $\sum\limits_{i=1}^{n} a_{1} < 1$ I have been trying to show that given $$P_{n}(\alpha) = \alpha^{n} - a_{1}\alpha_{n-1} - a_{2}\alpha^{n-2}... - a_{n} = 0,$$ the $\alpha$'s that solve this equation (real-valued or complex) lie in the unit circle only if $\sum\limits_{i=1}^{n} a_{i} < 1$. This result is referenced at the bottom of the reply in the following link Are all $AR(p)$ processes for which $|a_1|,....,|a_p| < 1$ stationary?.
By first assuming only real roots, I have argued as follows. Suppose all roots lie inside the unit circle, i.e. $|\alpha| < 1$ for every root $\alpha$. Then for all $x \geq 1$ we must have that $P_{n}(x) > 0$ or $P_{n}(x) < 0$. If not, then by the intermediate value theorem there would exist a root of magnitude larger than $1$. Since for large enough $x$ we must have $P_{n}(x) > 0$ (since $\lim\limits_{x\rightarrow \infty} P_{n}(x) = \infty$), we conclude that $P_{n}(x) > 0$ for all $x\geq 1$. It follows that $P_{n}(1) = 1 - \sum\limits_{i = 1}^{n} a_{i} > 0$, which is equivalent to the statement $\sum\limits_{i = 1}^{n} a_{i} < 1$.
I do not see how I can extend this argument to account for complex roots. I would appreciate any help, both by reference and by explanation/hint. Thanks in advance to all that took the time to read this post.

Let
$$P(z) = z^n - \sum_{i=0}^{n-1} a_i z^i.$$
Notice that
$$P(1) = 1^n - \sum_{i=0}^{n-1} a_i 1^i = 1 - \sum_{i=0}^{n-1} a_i.$$

*

*If all the roots are inside the unit circle, there are no roots on the real interval $[1, \infty).$ Because the values of $P(x)$ increase to $+\infty$ as the real number $X$ grows large, $P$ must be positive everywhere on $[1,\infty).$ In particular, $P(1) \gt 0,$ equivalent to $\sum_{i=0}^{n-1} a_i \lt 1.$


*When $\sum_{i=0}^{n-1} a_i \lt 1,$ $P(1) \gt 0.$  However, this does not imply all the roots of $P$ lie inside the unit circle. The foregoing analysis suggests a way to construct counterexamples: give $P$ a zero in the interval $(1,\infty).$ Since $P(1)\gt 0$ and eventually $P(x)\gt 0$ for large $x,$ $P$ must have at least two zeros in this interval.  The simplest possible counterexamples will be quadratic.  Consider, for instance, $P(x) = (x-2)^2 = x^2 - 4x + 4,$ for which $a_1=4$ and $a_0=-4.$ $P$ has no zeros inside the unit circle but the sum of the $a_i$ is less than $1.$
Consequently, $\sum a_i\lt 1$ is a necessary but not sufficient condition for all the roots of $P$ to lie within the unit circle.
