Consider the sequence of symmetric matrices with diagonal 2 and second-diagonal s $-1$, e.g. $$ M_4= \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0\\ 0 & -1 & 2 & -1\\ 0 & 0 & -1 & 2\\ \end{pmatrix} $$
I've found out that the characteristic polynomials are $$ \begin{cases} P_1(x)=2-x\\ P_2(x)=(2-x)^2-1\\ P_n(x) = (2-x)P_{n-1}(x)-P_{n-2}(x) \end{cases} $$
Or with a variable change $$ \begin{cases} Q_1(y)=y\\ Q_2(y)=y^2-1\\ Q_n(y) = y Q_{n-1}(y)-Q_{n-2}(y) \end{cases} $$
Looking at the first 8 $P_n$

I see that all eigenvalues are real (as for any symmetric matrix), they are between 0 and 4.
- How can I prove that all eigenvalues are between 0 and 4?
- Are these polynomials known (have a name)?
- How can I prove that the polynomial are sandwitched between $$ \frac{1}{x}+\frac{1}{4-x}\quad\text{and}\quad -\frac{1}{x}-\frac{1}{4-x} $$

The Chebyshev polynomials of the second kind satisfy the recurrence relation $$ \begin{cases} U_0(y) = 1 \\ U_1(y)=x\\ U_n(y) = 2y U_{n-1}(y)-U_{n-2}(y) \end{cases} $$ so that $Q_n(y) = U_n(y/2)$ and $P_n(x) = U_n(1-x/2)$.
The zeros of $U_n$ are $$ y_k = \cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n $$ in the range $(-1, 1)$, so that $$ x_k = 2 - 2\cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n $$ are the zeros of $P_n$ in the range $(0, 4)$.
Also for $|x| < 1$ $$ U_n(x) = \frac{\sin((n+1)\arccos(x))}{\sqrt{1-x^2}} $$ which implies $$ |U_n(x)| \le \frac{1}{\sqrt{1-x^2}} $$ and therefore $$ | P_n(x)| \le \frac{2}{\sqrt{x(4-x)}} \le \frac 1x + \frac{1}{4-x} $$ for $0 < x < 4$, the last estimate follows from the inequality between harmonic and geometric mean.