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. 



Do all the roots of the polynomial lie in the unit disk?
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](http://en.wikipedia.org/wiki/Gauss%E2%80%93Lucas_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.


