Prove positive definiteness I want to prove that the matrix
$$\begin{pmatrix}
1   &\cfrac{1}{2} &\cfrac{1}{3} &\cdots &\cfrac{1}{n} \\
\cfrac{1}{2} &\cfrac{1}{3} &\cfrac{1}{4} &\cdots &\cfrac{1}{n+1} \\
&\vdots &&\ddots &\vdots \\
\cfrac{1}{n} &\cfrac{1}{n+1} &\cfrac{1}{n+2} &\cdots &\cfrac{1}{2n-1}
\end{pmatrix}$$
is positive definite. Using mathematical induction, I only need to show that its determinant is positive. But I can't find the way out.
 A: Hint. Call your matrix $H$. Then $H=\left(\int_0^1 x^{i+j}dx\right)_{i,j=1,2,\ldots,n}=\int_0^1 v(x)v(x)^Td x$, where $v(x)=(1,x,x^2,\ldots,x^{n-1})^T$. Can you see that $H$ is positive semi-definite? Now, if $u^THu=0$ for some vector $u$, then $\int_0^1 u^Tv(x)v(x)^Tu dx=0$. Can you infer that $u=0$ (think about Vandermonde matrix)?
Remark. From a wider perspective, $H$ is an instance of matrices of the form $A=\left(\frac1{1+x_i+x_j}\right)_{i,j=1,2,\ldots,n}$ where $x_i\ge0$ for all $i$ (in your case, $x_i=i-1$), which is known (q313249) to be positive semidefinite. The same trick in the above hint can be used to show that $A$ is positive definite iff all $x_i$s are distinct.
Edit. Alternatively, $H$ is a Hilbert matrix. Every leading principal submatrix of $H$ is again a Hilbert matrix. So, by Sylvester's criterion, it suffices to prove that every Hilbert matrix has a positive determinant. As Gerry Myerson has pointed in his comment, mathematical induction may be useful in this case.
