How I find limit of $P_n$

Solution 1:

$a_{n+1}=(n+1)(a_n+1)=(n+1)a_n+(n+1)$, and so on, you can get that $$a_{n+1}=(n+1)!\left(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}\right).$$ If that,$$P_n=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!},$$ So $$\lim P_n=e.$$

Solution 2:

Hint: Rewrite $P_n$ as: $$P_n=\left(\frac{a_1+1}{a_1}\right)(\frac{a_2+1}{a_2})\cdots(\frac{a_n+1}{a_n})$$ $$P_n=\frac{1}{a_1}(\frac{a_1+1}{a_2})(\frac{a_2+1}{a_3})\cdots(\frac{a_{n-1}+1}{a_n})(a_n+1)$$ $$P_n=\frac{1}{a_1}(\frac{a_1+1}{a_2})(\frac{a_2+1}{a_3})\cdots(\frac{a_{n-1}+1}{a_n})(a_n+1)$$ $$P_n=\frac{1}{a_1}(\frac{1}{2})(\frac{1}{3})\cdots(\frac{1}{n-1})(a_n+1)$$

Finally, solve the recurrence relation to get $a_n$

Solution 3:

From the recurrence we get that $\frac{a_n}{n!}-\frac{a_{n-1}}{(n-1)!}=\frac{1}{(n-1)!}$, that yields $\lim_{n\to\infty}\frac{a_n}{n!}=e$. Then $$\lim_{n\to\infty}\prod_{k=1}^{n}\frac{a_k+1}{a_k}=\lim_{n\to\infty}\prod_{k=1}^{n}\frac{a_{k+1}}{(k+1)a_k}=\lim_{n\to\infty}\frac{a_{n+1}}{(n+1)!}=e $$