let $a_{1} = 0, a_{2} =1$ and $a_{n+2} = \frac{(n+2)a_{n+1} - a_{n}}{n+1}$. Prove that $\lim\limits_{n \to +\infty} a_{n} = e$

Question

let $a_{1} = 0$, $a_{2} =1$ and $a_{n+2} = \frac {(n+2)a_{n+1} - a_{n}} {(n+1)}$. Prove that $\lim a_{n} = e$

Knowing that I have proved that $a_{n+1} -a_{n} = 1/n!$, then I shall do what next?

Answer 1

Since you've proved that $a_{n+1}-a_n = \frac{1}{n!}$ you have that $$a_n = \frac{1}{1!}+ \ldots + \frac{1}{(n-1)!}$$ So $$\lim\limits_{n \to +\infty} a_n = \lim\limits_{n \to \infty} \left( \frac{1}{1!} + \ldots + \frac{1}{(n-1)!} \right) = \sum\limits_{n=1}^{+\infty} \frac{1}{n!} = e-1$$

Are you sure the correct answer is $e$? Maybe it was just $a_0 = 0$ and $a_1 = 1$.