Sequence of zeros of polynomials converges to $\pi$ Let $x_n\in\mathbb{R}$ be the zero closest to $\pi$ of polynomial:
$$
p_n(x):=\sum_{k=0}^n \frac{(-1)^k}{(2k)!}\left(\frac{z}{2}\right)^{2k}.
$$
I want to prove that

*

*$x_n\to \pi$ as $n\to\infty$,

*$\forall R\in\mathbb{R}: R^n|x_n-\pi|\to 0$.

For the first part, I tried to use Mean Value Theorem by showing that:
$$
|x_n-\pi|=\left|\frac{p_n(x_n)-p_n(\pi)}{p_n'(a_n)}\right|=\left|\frac{p_n(\pi)}{p'_n(a_n)}\right|
\to 0 \quad\text{as}\quad n\to\infty
$$
where $a_n$ is some real number between $x_n$ and $\pi$.
It is clear that $p_n(\pi)\to \cos(\frac{\pi}{2})=0$.
However, I couldn't show that $\lim_{n\to\infty}p'_n(a_n)\neq 0$ which would lead to the desired result.
I wonder what I've been missing.
Further, I have no idea how to start for the second part. Can you give me some hint?
 A: For every $n \ge 1$, set
$$S_n(x) = \sum_{k=0}^n (-1)^k \frac{x^{2k}}{(2k)!}.$$
One only needs to show that for every $n$, $S_n$ has at least one unique root in the interval $[\pi/4,3\pi/4]$. Call $r_n$ one of these roots.
Once you have proved this, by compactness of $[\pi/4,3\pi/4]$, it sufficient to show that the only limit point of the sequence $(r_n)_{n \ge 1}$ is $\pi/2$. Let $(r_{\phi(n)})_{n \ge 1}$ be any subsequence converging to some $\ell \in [\pi/4,3\pi/4]$. Since $(S_n)_{n \ge 1}$ converges uniformly on $[\pi/4,3\pi/4]$ to the function $\cos$, one has
$$\cos(\ell) = \lim_{n \to +\infty} S_{\phi(n)}(r_{\phi(n)}) = 0.$$
so $\ell=\pi/2$. This proves (1).
To prove $(2)$, observe that $(S'_n)_{n \ge 1}$ converges uniformly on $[\pi/4,3\pi/4]$ to the function $-\sin$, so
$$\min_{x \in [-\pi/4,\pi/4]} -S'_n(x) \to \min_{x \in [-\pi/4,\pi/4]} \sin(x) = \frac{\sqrt{2}}{2}.$$
Hence there exists some integer $N \ge 1$ such that
$$\forall n \ge N, \quad \min_{x \in [-\pi/4,\pi/4]} -S'_n(x) \ge \frac{1}{2}.$$
Then, when $n \ge N$  on the one hand
$$|S_n(\pi/2)| = |S_n(\pi/2) - S_n(r_n)| \ge \frac{1}{2}|\pi/2-r_n|.$$
On the other hand, since the sequence $((\pi/2)^{2n}/(2n)!)_{n \ge 1}$ decreases and converges to $0$, the criterion for alternating series yields
$$|S_n(\pi/2)| = |S_n(\pi/2) - \cos(\pi/2)| \le \frac{(\pi/2)^{2n+2}}{(2n+2)!}$$
Statement $(2)$ follows.
Now let us prove that $S_n$ has at least one root in the interval $[\pi/4,3\pi/4]$.
When $n=1$, $S_n(x) = 1-x^2/2$, so $\sqrt{2}$ is a root.
For $n \ge 2$, we use Taylor - Lagrange formula
$$\cos x = S_n(x) + R_n(x) \textrm{ where } R_n(x) = \cos^{(2n+2)}(\theta_x x) \frac{x^{2n+2}}{(2n+2)!} \textrm{ for some } \theta_x \in [0,1].$$
Since the function $\cos^{(2n+2)} =(-1)^{n+1}\cos$ takes values in $[-1,1]$, we have
$$\forall x \in [0,3\pi/4], \quad |R_n(x)| \le \frac{x^{2n+2}}{(2n+2)!} \le \frac{(3\pi/4)^{2n+2}}{(2n+2)!} < \frac{\sqrt{2}}{2} .$$
Since $\cos(\pi/4) = \cos(3\pi/4) = \sqrt{2}/2$, we derive $S_n(\pi/4) > 0 > S_n(3\pi/4)$ and the intermediate value theorem applies. The proof is complete.
