Compute the limit of a determinant of a given $n \times n$ matrix Compute:
$\displaystyle \lim_{n \rightarrow \infty} \begin{vmatrix} 
1+x & -x & 0 & 0 & \cdots & 0 & 0 \\
-\frac{1}{2} & 1+\frac{x}{2} & -x & 0 & \cdots & 0 & 0 \\
0 & -\frac{1}{3} & 1+\frac{x}{3} & -x & \cdots & 0 & 0 \\
0 & 0 & -\frac{1}{4} & 1+\frac{x}{4} & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 1+\frac{x}{n-1} & -x \\
0 & 0 & 0 & 0 & \cdots & -\frac{1}{n} & 1+\frac{x}{n} \\
\end{vmatrix}$
Let $D_n$ be the given determinant. I solved it via mathematical induction and got the following solution:
$\displaystyle \lim_{n \rightarrow \infty} D_n = \lim_{n \rightarrow \infty} \sum_{i=0}^n \frac{x^i}{i!} = \sum_{i=0}^{\infty} \frac{x^i}{i!} = \mathrm{e}^x$
I want to solve it by using recurrence relations:
We know that $D_1 = 1+x$ and $D_2 = 1+x+\frac{x^2}{2}$. If I develop the determinant $D_n$ by the last column I get:
$D_{n+2} = \left( 1+\frac{x}{n+2} \right) D_{n+1} - \left( \frac{x}{n+2} \right) D_{n} \text{ if } n \geq 1$.
How can I solve this linear recurrence relation with variable coefficients?

Hint : $D_{n+2} = \left( 1+\dfrac{x}{n+2} \right) D_{n+1} - \left( \dfrac{x}{n+2} \right) D_{n}$ rewrites as $$D_{n+2} -D_{n+1} = \frac{x}{n+2} (D_{n+1} - D_{n})$$
Let $A_n = D_{n+1}-D_n$ : you get $$A_{n+1} = \dfrac{x}{n+2}A_n$$
Can you proceed from here ?
