Prove the zeros of a polynomial all lie in an annulus. I am working on a problem. It has two parts:

(a) Let $c_{0}>c_{1}>\cdots c_{n}>0$. Show that the polynomial $P(z):=c_{0}+c_{1}z+\cdots+c_{n}z^{n}$ has no zeros inside the closed unit disc. 
(b) Show that the zeros of polynomial $P_{n}(z):=1+\frac{z}{2}+\frac{z^{2}}{3}+\cdots+\frac{z^{n}}{n+1}$ all lie in an annulus $\{1<|z|<1+\delta_{n}\}$ where $\delta_{n}\rightarrow 0$ as $n\rightarrow\infty$.

I've proved part (a), but I am stuck in part (b). I think the proof of part (b) may be similar to part (a), so I state part (a) above and give my proof below, then I will give my attempt for part (b).
Part (a):
Suppose there exists $z_{0}\in\mathbb{C}$ such that $P(z_{0})=0$ and $|z_{0}|<1$.
Since $P(z_{0})=0$, we also have $$(1-z_{0})P(z_{0})=0, $$ where $LHS=c_{0}+(c_{1}-c_{0})z_{0}+(c_{2}-c_{1})z_{0}^{2}+\cdots+ (c_{n}-c_{n-1})z_{0}^{n}-c_{n}z_{0}^{n+1}.$
Thus, we have $$c_{0}=(c_{0}-c_{1})z_{0}+(c_{1}-c_{2})z_{0}^{2}+\cdots (c_{n-1}-c_{n})z_{0}^{n}+c_{n}z_{0}^{n+1}.$$
Now, taking norm to both side, and recalling that $c_{0}>c_{1}>\cdots>c_{n}>0$ and $|z_{0}|<1$, we have 
\begin{align*}
c_{0}&<c_{0}-c_{1}+c_{1}-c_{2}+\cdots+c_{n-1}-c_{n}+c_{n}\\
&=c_{0}
\end{align*}
which is a contradiction. 
Thus, there is no zero of $P(z)$ that is inside the closed unit disc. 
Part (b):
For part (b), I mimic what I've done in part (a). Let $z\in\mathbb{C}$ be a zero of $P_{n}(z)$, then $P_{n}(z)=0$ implies that $$(1-z)P_{n}(z)=0.$$
Thus, we have $$\Big(1-\dfrac{z^{n+1}}{n+1}\Big)-\dfrac{z}{2}-\dfrac{z^{2}}{6}-\cdots-\dfrac{z^{n}}{n(n+1)}=0.$$
Set $Q_{n}(z):=-\dfrac{z}{2}-\dfrac{z^{2}}{6}-\cdots-\dfrac{z^{n}}{n(n+1)}.$
Then if $|z|<1$, we have 
\begin{align*}
1&\leq|Q_{n}(z)|+\Big|\dfrac{z^{n+1}}{n+1}\Big|\\
&\leq\dfrac{|z|}{2}+\dfrac{|z|^{2}}{6}+\cdots+\dfrac{|z|^{n}}{n(n+1)}+\dfrac{|z|^{n+1}}{n+1}\\
&<\dfrac{1}{2}+\dfrac{1}{6}+\cdots+\dfrac{1}{n(n+1)}+\dfrac{1}{n+1}\\
&=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots+\dfrac{1}{n}-\dfrac{1}{n+1}+\dfrac{1}{n+1}\\
&=1,
\end{align*}
which is a contradiction. 
Thus, zeros of $P_{n}(z)$ must lie in $|z|\geq 1$.
Then, I try to get rid of $|z|=1$ by using the same techniques, but I found something else interesting. 
If $|z|=1$, then by definition
$$|Q_{n}(z)|\leq 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n}{n+1}.$$
On the other hand, since we assume $z$ is a zero, we have $$|Q_{n}(z)|=\Big|1-\dfrac{z^{n+1}}{n+1}\Big|\geq \Big|1-\dfrac{1}{n+1}\Big|=\dfrac{n}{n+1}.$$
Thus, if $|z|=1$, we have $$\dfrac{n}{n+1}\leq |Q_{n}(z)|\leq\dfrac{n}{n+1},$$ and thus $$|Q_{n}(z)|=\dfrac{n}{n+1}.$$
This did not give me any contradiction, but an idea of $\delta_{n}$. By the problem itself, we can see that if $\delta_{n}\rightarrow 0$, then the annulus will become to a unit circle $|z|=1$, which is exactly our case. 
Also, by my argument above, I want my $\delta_{n}$ to be related to $|Q_{n}(z)|$, so I tried to set $$1+\delta_{n}=\dfrac{n}{n+1},$$ which gives us $$\delta_{n}=-\dfrac{1}{n+1},$$ which tends to be $0$ as $n\rightarrow\infty$. 
But then I don't know how to proceed, what should I do now?
Thank you!
 A: Consider the polynomial $R_n(z)=(z-1)P_{n}(z)=-1+\dfrac{z}{2}+\dfrac{z^{2}}{6}+\cdots+\dfrac{z^{n}}{n(n+1)}+\dfrac{z^{n+1}}{n+1}$. 
We will show that the Cauchy bound $\rho_n=1+\gamma_n$ of $R_n$, satisfies $\gamma_n \to 0$ as $n \to \infty$ hence we are done since all the roots $w_{k,n}$ of $R_n$ (hence of $P_n$) satisfy $|w_{k,n}| \le \rho_n$
If not familiar with the notion, the Cauchy bound $\rho_P$ of a polynomial $P(z)=\sum_{0}^{n}{a_kz^k}, a_n \ne 0, n \ge 1$ is the unique positive root of the polynomial $|a_n|z^n-\sum_{0}^{n-1}|a_k|z^k$ - taken by convention as $0$ if the polynomial is just $a_nz^n$ -and by the triangle inequality the relation $|w_k| \le 
\rho_P$ for all the roots of $P$ is clear. 
Also it is clear that if for some $R >0, \sum_{k=0}^{n-1}{|a_k|R^k} \le |a_n|R^n$, then $\rho_P \le R$ by the unicity of the positive root.
We use the inequality: $\rho_P \le max_{k=0,..,n-1}|n\frac{a_k}{a_n}|^{\frac{1}{n-k}}$, which is fairly obvious since if we denote by $R$ the maximum on the RHS, $|a_k| \le \frac{1}{n}|a_n|R^{n-k}, k=0,...n-1$, so $\sum_{k=0}^{n-1}{|a_k|R^k} \le |a_n|R^n$, hence $R \ge \rho_P$ as above.
In our case the degree is $n+1$ and the coefficients are $|a_0|=1, a_k=\frac{1}{k(k+1)}, 1 \le k \le n, a_{n+1}=\frac{1}{n+1}$, so we need to prove the following two results:
1: $((n+1)^2)^{\frac{1}{n+1}}=1+b_n, b_n \to 0$
2: $max_{ 1 \le k \le n}(\frac{(n+1)^2}{k(k+1)})^{\frac{1}{n+1-k}}=1+c_{n}, c_{n} \to 0$
But using the inequality $\log(1+x) \ge \frac{x}{2}, 0 \le x \le 1$, we get $\log(1+b_n) = 2\frac{\log(n+1)}{n+1} \to 0$, hence $b_n \le 1$ eventually, hence $b_n \le 4\frac{\log(n+1)}{n+1} \to 0$, so 1: is done.
Similarly if $k \le \frac{2n}{3}$, $n+1-k \ge \frac{n}{3}$ and the same proof applies since then $\frac{\log(\frac{(n+1)^2}{k(k+1)})}{n+1-k} \le 6\frac{\log(n+1)}{n} \to 0$
If $k \ge \frac{2n}{3}, \frac{\log(\frac{n+1}{k})}{n+1-k} \le \frac{n+1-k}{k(n+1-k)} \le \frac{2}{n} \to 0$ and same for the other $\frac{\log(\frac{n+1}{k+1})}{n+1-k} \to 0$, so we are done too for case 2; and the problem is solved with $\delta_n$ the maximum of $b_n, c_n$ above
(we used $\log(\frac{n+1}{k})=\log(1+\frac{n+1-k}{k}) \le \frac{n+1-k}{k}$)
