TITLE: $p(z) = 1 + 2z + 3z^2 + \dots + (n+1)z^n$ has no zeroes in a disk for sufficiently large $n$.
QUESTION [5 upvotes]: Let $0 < r < 1$.  I need to prove that $p(z) = 1 + 2z + 3z^2 + \dots +(n+1)z^n$ has no zeroes in the disk $|z| < r$ if $n$ is sufficiently large.  I'm thinking Rouche's theorem might be helpful, but I'm not exactly sure how to apply it. Help would be appreciated.

REPLY [7 votes]: Let $p_n(z) = \sum_{k=0}^n (k+1)z^k$, and $p(z) = \frac{1}{(1-z)^2}$. Note that $p$ is analytic on $B(0,1)$ and has no zeros or poles in $B(0,1)$.
Find a lower bound for $|p(z)|$ on $B(0,1)$.
Hint:

 If $|z|<1$, we have  $|1-z| < 2$, hence $|p(z)|> \frac{1}{4}$.

Choose $r \in (0,1)$, and $r < s <1$. Then show that $p_n \to p$ uniformly on $B(0,s)$. Find a suitable upper bound for $|p_n(z) -p(z)|$ in $B(0,s)$.
Hint:

 Now choose $N$ such that if $n \ge N$, we have $|p_n(z) -p(z)| \le \frac{1}{4}$ for all $z \in B(0,s)$.

Finally, apply Rouché's theorem on $|z|=r$.

 Hence if $|z|=r$, we have $|p_n(z) -p(z)| \le \frac{1}{4} < |p(z)|$. We conclude from Rouché's theorem that $p_n$ the same number of zeros as $p$ in $B(0,r)$ (that is, zero zeros).