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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.

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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.

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  • $\begingroup$ Neat and clean! +1 $\endgroup$ – user1551 Apr 27 '13 at 8:45
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    $\begingroup$ Indeed, it's easy to see that the rank of $H_n$ equals $1$, so that $0$ is an eigenvalue of $H_n$ of multiplicity $(n-1)-1$; therefore the eigenvalues of $I_{n-1}+H_n$ are $1$ with multiplicity $n-2$ and $h_n$ with multiplicity $1$, reproving the exact formula. $\endgroup$ – Greg Martin Apr 28 '13 at 2:30
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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.

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