limit of the determinant of $n\times n$ matrix over $n!$ Let $n$ be a positive integer, and let $d_n$ denote the determinant of the following $n\times n$ matrix
$M=\begin{bmatrix}
   2       & 1 & 1&1 & \dots & 1 \\
   1       & 3 & 1&1 & \dots &1 \\
   1 &1&4&1 &\dots &1  \\
    1       & 1&1 & \ddots & 1 & 1\\1&1 &\dots&\dots&n &1\\ 1&1 &\dots&\dots&\dots &n+1
\end{bmatrix}
$
What is $\lim_{n\rightarrow \infty} \frac{d_n}{n!}$?
For $n=1, \det(M)=2,  \frac{d_1}{1!}=2$.
For $n=2, \det(M)=5,  \frac{d_2}{2!}=\frac{5}{2}$.
For $n=3,\det(M)=17,  \frac{d_3}{3!}=\frac{17}{6} $.
It looks like the sequence is increasing. But figuring out $d_n$ is tricky. I'm not sure if https://en.wikipedia.org/wiki/Determinant#n.C2.A0.C3.97.C2.A0n_matrices will help.
Thanks in advance!
 A: You can define $A$ as $n\times n$ matrix with all entries $1$ and $B=diag(1,2,...n)$ and then $M=A+B$. Then
we can use the determinant of sum of two matrix formula as in the following result:
Let $A$ and $B$ be two $n\times n$ square matrix on any field, then
$$\det(A+B)=\sum\limits_{\alpha ,\beta } {{{( - 1)}^{s(\alpha ) + s(\beta )}}\det A[\alpha |\beta ]\det B(\alpha |\beta )} $$
where the summation ranges for $\alpha ,\beta  \subseteq {\rm{\{ }}1,2,....,n\}$ satisfying Card($\alpha$)=Card($\beta$) and  $s:P(\{ 1,2,...,n\} ) \to {N }$ is defined as the sum of all elements in a subset of $\{1,2,...,n\}$(by convention , it maps empty set to zero),$P$ means the power set. Note here that $A[\alpha|\beta]$ means the matrix block of $A$ with rows $\alpha$ and columns $\beta$ and $B(\alpha|\beta)$ means the matrix block of $B$ excluding rows of $\alpha$ and columns of $\beta$.
Then we can see by $A,B$, the only non-zero $\det A[\alpha|\beta]\det B(\alpha|\beta)$ is when $\alpha=\beta=\emptyset$ or when $\alpha=\beta$ being a singleton. Then we can find that when $\alpha=\beta=\emptyset$, the summand is $\det(B)=n!$ and when $\alpha=\beta=k=1,2,...,n$, the $\det A[\alpha|\beta]=1$ and the $\det B(\alpha|\beta)=n(n-1)...(k-1)(k+1)...1$ and thus we see det(M)=$n!+\sum_{k=1,2,...,n} n!/k$ and thus we see $\det(M)/n!=1+\sum_{k=1,2,...n}1/k=1+1+1/2+1/3+...+1/n$ is unbounded when $n$ goes to infinity!!!
