# Proving that roots of $P_n(z) := \sum k^3 z^k$ satisfy $|z| < 1$.

For each positive integer $n$, define the polynomial $$P_n(z) = z + 2^3z^2 + ... + n^3z^n.$$ Prove that the roots of $P_n(z)$ lie inside the unit circle.

Assuming the existence of $|\alpha| \geq 1, P_n(\alpha) = 0$, I tried using the triangle inequality: $$|\alpha|^nn^3 = |\alpha + 8\alpha^3 + ... + (n-1)^3 \alpha^{n-1} | \leq |\alpha| + ... + (n-1)^3 |\alpha|^{n-1}$$ which seems to be too weak to create a contradiction.

Thoughts/solutions are appreciated.

• have you tried rouche's theorem? Sep 5 '17 at 2:49

The roots of $f_n(z) = 1 + z + \ldots + z^n = (z^n-1)/(z-1)$ are roots of unity, and lie on the unit circle. The roots of $f'_n(z) = \sum_{j=1}^n j z^{j-1}$ lie in the convex hull of these. Since the roots of $f_n$ are all simple, they are not roots of $f'_n$, so the roots of $f'_n$ are (strictly) inside the unit circle.
The roots of $g_n(z) = z f'_n(z) = \sum_{j=1}^n j z^j$ are $0$ and the roots of $f'_n(z)$, so again these are strictly inside the unit circle.
Iterating this process, we find that for all positive integers $k$, the roots of $\sum_{j=1}^n j^k z^j$ are strictly inside the unit circle.