<s>[INST] 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.
 [/INST] 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.
</s>