Is this determinant bounded? Let $D_n$ be the determinant of the $n-1$ by $n-1$ matrix such that the main diagonal entries are $3,4,5,\cdots,n+1$ and other entries being $1$. i.e.
$$D_n= \det \begin{pmatrix}
3&1&1&\cdots&1\\
1&4&1&\cdots&1\\
1&1&5&\cdots&1\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&1&1&\cdots &n+1
\end{pmatrix}$$
Is the set $\{D_n/n! \ | \ n=2,3,4,\cdots\}$ bounded?
My attempt is to observe that $D_n= \det ((1)+ diag\{2,3,\cdots,n\})$, where $(1)$ is the matrix with all entries being $1$. We have $\det (1)=0$ and $\det (diag\{2,3,\cdots,n\})=n!$, which seems to be promising, but then I can't move along, could someone please helps.
 A: Since adding a multiple of one row to another doesn't change a matrix's determinant, we can subtract the first row from each of the other rows to see that
$$D_n= \det \begin{pmatrix}
3&1&1&\cdots&1\\
-2&3&0&\cdots&0\\
-2&0&4&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-2&0&0&\cdots &n
\end{pmatrix}.$$
Then we subtract $1/(k+1)$ times the $k$th row from the first row for each $2\le k\le n-1$, yielding
$$D_n= \det \begin{pmatrix}
A&0&0&\cdots&0\\
-2&3&0&\cdots&0\\
-2&0&4&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-2&0&0&\cdots &n
\end{pmatrix}$$
with $A = 3 + \frac23 + \frac24 + \cdots + \frac2n = 2h_n$, where $h_n = \sum_{k=1}^n \frac1k$ is the $k$th harmonic number. Since the determinant of an upper triangular matrix is just the product of its diagonal entries, this computation shows that
$$
D_n = h_nn!.
$$
In particular, $D_n/n!$ is unbounded.
A: Let $H_n= \left( \begin{array}{cccc} 1/2 & 1/3 & \dots & 1/n \\ \vdots & \vdots & & \vdots \\ 1/2 & 1/3 & \dots & 1/n \end{array} \right)$. Then, $\displaystyle \frac{D_n}{n!}=\det(I_{n-1}+H_n)$. Noticing that $H_n= \left( \begin{array}{c} 1/2 \\ 1/3 \\ \vdots \\ 1/n \end{array} \right) \left( \begin{array}{ccc} 1 & \dots & 1 \end{array} \right)$, you deduce that $\displaystyle H_n^2= \left( \sum\limits_{k=2}^n \frac{1}{k} \right)H_n$. Moreover, you find that $X^2-(1+h_n)X+h_n$ is the minimal polynomial of $I_{n-1}+H_n$, with $\displaystyle h_n= \sum\limits_{k=1}^n \frac{1}{n}$ the $n$th harmonic number. Therefore, the eigenvalues of $I_{n-1}+H_n$ are exactly $1$ and $h_n$, hence $\det(I_{n-1}+H_n) \geq h_n$. (In fact, Greg Martin showed there is equality.)
Consequently, the sequence $(D_n/n!)$ is not bounded.
