For $n \geq 2,$ consider the following square matrix of order $(n-1)$
\begin{array}{cccccc} 3 & 1 & 1 & 1 & & 1 \\ 1 & 4 & 1 & 1 & \dots & 1 \\ 1 & 1 & 5 & 1 & \cdots & 1 \\ 1 & 1 & 1 & 6 & & 1 \\ & & & & \ddots & \vdots \\ 1 & 1 & 1 & 1 & \cdots & n+1 \end{array}
Find its determinant using only elementary row operations and denote it by $A_{n}$. Hence or otherwise, check whether the sequence $\left\{^{A_{n}} / n_{1}\right\}_{n>2}$ is bounded
Basically I am getting no clue. Should I proceed with Echelon Form?
Basic idea:
1. We get rid of $1$s except for the first row
2. We get rid of $1$s in the first row
3. We consider $(A)_{1,1}$ as the rest is known
So 1. we subtract the 1st row from every other row: $$\begin{align*} \left|\begin{array}{cccccc} 3 & 1 & 1 & 1 & & 1 \\ 1 & 4 & 1 & 1 & \dots & 1 \\ 1 & 1 & 5 & 1 & \cdots & 1 \\ 1 & 1 & 1 & 6 & & 1 \\ & & & & \ddots & \vdots \\ 1 & 1 & 1 & 1 & \cdots & n+1 \end{array}\right| &= \left|\begin{array}{cccccc} 3 & 1 & 1 & 1 & & 1 \\ -2 & 3 & 0 & 0 & \dots & 0 \\ -2 & 0 & 4 & 0 & \cdots & 0 \\ -2 & 0 & 0 & 5 & & 0 \\ & & & & \ddots & \vdots \\ -2 & 0 & 0 & 0 & \cdots & n \end{array}\right| \end{align*}$$ 2. we subtract each row (say $k$th), multiplied by $\frac{1}{k+1}$ from the first row: $$= \left|\begin{array}{cccccc} a_{1,1} & 0 & 0 & 0 & & 0 \\ -2 & 3 & 0 & 0 & \dots & 0 \\ -2 & 0 & 4 & 0 & \cdots & 0 \\ -2 & 0 & 0 & 5 & & 0 \\ & & & & \ddots & \vdots \\ -2 & 0 & 0 & 0 & \cdots & n \end{array}\right|$$ 3. As it's lower-triangular now, we consider $a_{1,1}$ as $|A|=a_{1,1}\prod\limits_{k=3}^n k$,
$a_{1,1}=3+2\sum\limits_{k=3}^n \frac{1}{k}$