Calculate the determinant of the matrices $a_{ij}=\frac{1}{i+j-1}$ and $b_{ij}=\frac{1}{i+j}$?

Question

I would like to know if there is any formula for calculating determinants of the following symmetric matrices: $$ A=[a_{ij}]_{n\times n},\qquad a_{ij}=\frac{1}{i+j-1}, $$ and $$ B=[b_{ij}]_{n\times n},\qquad b_{ij}=\frac{1}{i+j},\qquad $$ where $1\leq i,j\leq n$. If there are no formula for the determinants, is it true that the determinant of these matrices are never zero.

Thanks!

The first one is the Hilbert matrix, Hilbert has worked out a closed form for its determinant, see the wiki page. I think I have seen the second matrix somewhere but can't remember where it is. — achille hui, Dec 18 '15 at 19:27

score 3 · Accepted Answer · answered Dec 18 '15 at 19:29

For your $a_{ij}$ matrix, the answer is given by the reciprocal of this sequence in the OEIS (so in the $n=1$ case the determinant is 1, in the $n=2$ case it's 1/12, in the $n=3$ case it's 1/2160, and so on.) For your $b_{ij}$ matrix, the answer is given by the reciprocal of this sequence.) According to the OEIS the formulas you want are:

$$\det A_n = {\prod_{k=0}^{n-1} k!^2 \over n^n \prod_{k=1}^{n-1} (n^2-k^2)^{n-k}}$$ and $\det B_n = \det A_n / {2n \choose n}$.

A note in the OEIS says that these formulas are proven from results on Cauchy matrices.

Calculate the determinant of the matrices $a_{ij}=\frac{1}{i+j-1}$ and $b_{ij}=\frac{1}{i+j}$?

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