How to prove that the following matrix has an inverse with integer entries? Also find out the inverse.
\begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} & \cdots &\ \frac{1}{n-1} & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \cdots &\ \frac{1}{n} & \frac{1}{n+1} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \cdots &\ \frac{1}{n+1} & \frac{1}{n+2} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \cdots &\ \frac{1}{n+1} & \frac{1}{n+2} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ \frac{1}{n} & \frac{1}{n+1} & \frac{1}{n+2} & \cdots &\ \frac{1}{2n-2} & \frac{1}{2n-1} \\ \end{bmatrix}
I am assuming that the question is about the matrix $A$ whose $(i,j)$ entry is $A_{ij}=\frac{1}{i+j-1}$. (As remarked in the comments, there is a typo as written in the question.)
I claim that $A^{-1}$ has the following explicit formula: $$ \left(A^{-1}\right)_{ij}=(-1)^{i+j}\binom{n+i-1}{n-j}\binom{n+j-1}{n-i}\binom{i+j-2}{i-1}^2(i+j-1). $$
To verify that this is indeed the inverse, I would suggest using induction to show that its product with $A$ is the identity matrix.