Prove the matrix is positive Consider the matrix $A=\begin{bmatrix} 
    1 & 1/2 & 1/3 &\dots &1/n \\
    1/2 & 1/3 & 1/4 &\dots  &1/(n+1) \\
    \vdots & \vdots & \vdots & \vdots & \vdots\\
    1/n & 1/(n+1) & 1/(n+2) & \dots& 1/(2n-1)
    \end{bmatrix}$
Prove that $A$ is positive.
My work: $A$ is diagonalisable, symmetric but I can't seem to put these facts togheter to help me. I tried to prove by induction (a naive attempt) that the determinant of its minors is always positive but knowing $det(A^{k,k})>0$ there is no information of $det(A^{k+1,k+1}).

tried two , Sylvester Inertia
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 \frac{ 1 }{ 2 }  & 1 & 0 \\ 
 \frac{ 1 }{ 3 }  & 1 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
60 & 0 & 0 \\ 
0 & 5 & 0 \\ 
0 & 0 &  \frac{ 1 }{ 3 }  \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  \frac{ 1 }{ 2 }  &  \frac{ 1 }{ 3 }  \\ 
0 & 1 & 1 \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
60 & 30 & 20 \\ 
30 & 20 & 15 \\ 
20 & 15 & 12 \\ 
\end{array}
\right) 
$$
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrrr} 
1 & 0 & 0 & 0 \\ 
 \frac{ 1 }{ 2 }  & 1 & 0 & 0 \\ 
 \frac{ 1 }{ 3 }  & 1 & 1 & 0 \\ 
 \frac{ 1 }{ 4 }  &  \frac{ 9 }{ 10 }  &  \frac{ 3 }{ 2 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
420 & 0 & 0 & 0 \\ 
0 & 35 & 0 & 0 \\ 
0 & 0 &  \frac{ 7 }{ 3 }  & 0 \\ 
0 & 0 & 0 &  \frac{ 3 }{ 20 }  \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
1 &  \frac{ 1 }{ 2 }  &  \frac{ 1 }{ 3 }  &  \frac{ 1 }{ 4 }  \\ 
0 & 1 & 1 &  \frac{ 9 }{ 10 }  \\ 
0 & 0 & 1 &  \frac{ 3 }{ 2 }  \\ 
0 & 0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrrr} 
420 & 210 & 140 & 105 \\ 
210 & 140 & 105 & 84 \\ 
140 & 105 & 84 & 70 \\ 
105 & 84 & 70 & 60 \\ 
\end{array}
\right) 
$$
