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}).
 A: tried two , Sylvester Inertia
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 \frac{ 1 }{ 2 }  & 1 & 0 \\ 
 \frac{ 1 }{ 3 }  & 1 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
60 & 0 & 0 \\ 
0 & 5 & 0 \\ 
0 & 0 &  \frac{ 1 }{ 3 }  \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  \frac{ 1 }{ 2 }  &  \frac{ 1 }{ 3 }  \\ 
0 & 1 & 1 \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
60 & 30 & 20 \\ 
30 & 20 & 15 \\ 
20 & 15 & 12 \\ 
\end{array}
\right) 
$$
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrrr} 
1 & 0 & 0 & 0 \\ 
 \frac{ 1 }{ 2 }  & 1 & 0 & 0 \\ 
 \frac{ 1 }{ 3 }  & 1 & 1 & 0 \\ 
 \frac{ 1 }{ 4 }  &  \frac{ 9 }{ 10 }  &  \frac{ 3 }{ 2 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
420 & 0 & 0 & 0 \\ 
0 & 35 & 0 & 0 \\ 
0 & 0 &  \frac{ 7 }{ 3 }  & 0 \\ 
0 & 0 & 0 &  \frac{ 3 }{ 20 }  \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
1 &  \frac{ 1 }{ 2 }  &  \frac{ 1 }{ 3 }  &  \frac{ 1 }{ 4 }  \\ 
0 & 1 & 1 &  \frac{ 9 }{ 10 }  \\ 
0 & 0 & 1 &  \frac{ 3 }{ 2 }  \\ 
0 & 0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrrr} 
420 & 210 & 140 & 105 \\ 
210 & 140 & 105 & 84 \\ 
140 & 105 & 84 & 70 \\ 
105 & 84 & 70 & 60 \\ 
\end{array}
\right) 
$$
A: (Migrated from comment)
The matrix in question is called the Hilbert matrix. To see that $A$ is positive-definite, let $x \in \mathbb{R}^n$. Then
$$ x^T A x
= \sum_{i, j = 1}^{n} \frac{x_i x_j}{i+j-1}
= \sum_{i, j = 1}^{n} \int_{0}^{1} t^{i+j-2} x_i x_j \, dt
= \int_{0}^{1} \left( \sum_{i=1}^{n} x_i t^{i-1} \right)^2 \, dt
\geq 0 $$
Moreover, the equality in the last step holds if and only if $\sum_{i=1}^{n} x_i t^{i-1} \equiv 0$ on $t \in [0, 1]$, which is equivalent to $x = 0$. Therefore $A$ is positive-definite as required.
A: This is not a new answer. It is essentially the same ideas as already presented by  Sangchool Lee. I'm giving it just as yet another application of the identity 
$$ \frac{1}{x} = \int_0^\infty e^{-sx} ds.$$ 
Observe:  
$A_{i,j} = \frac{1}{i+j-1}$. 
\begin{align*} (Av,v) &= \sum_{i,j} \frac{v_j v_i}{i+j-1}\\
  & = \sum_{i,j} \int_0^\infty  (e^{-(j+1/2)s} v_j) (e^{-(i+1/2)s}v_i) ds \\
  & = \int_0^\infty \sum_{i,j}   (e^{-(j+1/2)s} v_j) (e^{-(i+1/2)s}v_i)ds \\
  & = \int_0^\infty \sum_{j} (e^{-(j+1/2)s} v_j)^2 ds\\
  & \ge 0
\end{align*} 
Note that the proof works whenever $A_{i,j} = 1/(f(i) + f(j))$ where $f$ is a positive function. 
