Do all the roots of the polynomial lie in the unit disk? How to prove (or to disprove) that all the roots of the polynomial of degree $n$ $$\sum_{k=0}^{k=n}(2k+1)x^k$$ belong to the disk $\{z:|z|<1\}?$ Numerical calculations confirm that, but I don't see any approach to a proof of so simply formulated statement. It would be useful in connection with an irreducibility problem. 
 A: The idea is taken from this other question Polynomial with the primes as coefficients irreducible?
Show instead that $f(1/x)$ has all roots lying outside of the unit disk. For that, multiply by $(x-1)$ and equate to $0$ obtaining:
$$x^{k+1}+\sum_1^k 2x^j=2k+1$$
Take absolute values and apply the triangular inequality and one obtains:
$$|x^{k+1}|+\sum_1^k 2|x^j|\geq \left|x^{k+1}+\sum_1^k 2x^j\right|=2k+1$$
This is clearly non possible if 
$|x|<1$. Moreover, if $|x|=1$ there is an equality which means that all the terms are aligned. In particular $2x^2/2x$ is real, so the only possibility is $x=1,-1$. But neither is a root of $f(1/x)$ so you are done.  
A: Let $f$ denote your polynomial. Roots of the polynomial
$$g(x)=\sum_{k=0}^nx^{2k+1}=x\frac{x^{2(n+1)}-1}{x^2-1}$$
are $0$ and roots of unity. By the Gauss–Lucas theorem, the roots of $g'(x)=f(x^2)$ lie in their convex hull, and a fortiori in the disk $\{z:|z|\le1\}$. In order to get a strict inequality, it suffices to show that $g$ is square-free.
A: The standard approach to this type of question is to use the Schur-Cohn procedure.
