How to prove the positive-definiteness of a symmetric Toeplitz matrix like this? Define a symmetric Toeplitz matrix by
$$\begin{pmatrix}c_1 & c_2 & c_3 & \cdots & c_n\\c_2 & c_1 & c_2 & \cdots & c_{n-1}\\c_3 & c_2 & c_1 & \cdots &c_{n-2}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\c_n & c_{n-1} & c_{n-2} & \cdots & c_1\end{pmatrix},$$ 
for $$c_1 = 1, \quad c_{k+1}=\frac{1}{2}\left((k+1)^{2-\alpha}-2 k^{2-\alpha}+(k-1)^{2-\alpha}\right) \quad (k>0),$$ where $\alpha \in (0, 1)$. 
How to prove that this matrix is positive-definite? Thanks a lot.
 A: Let $B := \{ B_{t} \}_{t\geq 0} $ be a fractional Brownian motion of Hurst parameter $1-\alpha/2 $ on some probability space $(\Omega,\mathcal{F},P)$. Since $B$ has the covariance function
$$
E(B_{t}B_{s}) = \frac{1}{2}\left( s^{2(1-\alpha/2)} + t^{2(1-\alpha/2)} - |t-s|^{2(1-\alpha/2)} \right),
$$ 
a simple calculation yields 
$$
E( (B_{n+k}-B_{n+k-1})(B_{n}-B_{n-1}) ) = \frac{1}{2}\left( (k+1)^{2-\alpha} + |k-1|^{2-\alpha} - 2k^{2-\alpha} \right)
$$
for any integers $n\geq 1$ and $ k\geq 0$ ($E$ means the expectation under $P$). Let us set (note that $B_{0}=0$ a.s.)
$$
 X=\left[ \begin{array}{c}
B_{1} \\
B_{2} \\
\cdots \\
B_{k}
 \end{array}\right],\ 
\Delta X=\left[ \begin{array}{c}
B_{1}-B_{0} \\
B_{2}-B_{1} \\
\cdots \\
B_{k}-B_{k-1}
 \end{array}\right],\ 
A=\left[ \begin{array}{c}
1 & 0 & 0 & \cdots & 0 \\
-1 & 1 & 0 & \cdots & 0 \\
0 & \ddots & \ddots & \ddots & \vdots \\
\vdots & \ddots &\ddots & \ddots & 0 \\
0 & \cdots &0 & -1 & 1  
 \end{array}\right],
$$
then the symmetric Toeplitz matrix of interest (which would be denoted by $C$) can be written as
$$
C = E\left( \Delta X(\Delta X)^{\top} \right)= A E(XX^{\top})A^{\top},
$$
where the superscript $\top$ means transpose.
Since $E(XX^{\top})$ is positive-definite (a proof of this fact can be found in, e.g., proposition 1.6 of "Selected Aspects of Fractional Brownian Motion" by Ivan Nourdin), $C$ is also positive-definite.
