Let $a_1=1$ and $a_n=n(a_{n-1}+1)$ for $n=2,3,...$.

Question

Let $a_1=1$ and $a_n=n(a_{n-1}+1)$ for $n=2,3,...$. Define $$P_n=\left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)...\left(1+\frac{1}{a_n}\right).$$ Find $\lim\limits_{n\to\infty} P_n$.

My approach: $$P_n=\left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)...\left(1+\frac{1}{a_n}\right)$$ $$=\left\{\frac{(a_1+1)(a_2+1)...(a_n+1)}{a_1.a_2...a_n}\right\}$$ $$=\left\{\frac{1}{a_1}\left(\frac{a_1+1}{a_2}\right)\left(\frac{a_2+1}{a_3}\right)...\left(\frac{a_{n-1}+1}{a_n}\right)(1+a_n)\right\}$$ $$=\left\{\frac{1}{a_1}.\frac{1}{2}.\frac{1}{3}.....\frac{1}{n}(1+a_n)\right\}$$ $$=\frac{1+a_n}{n!}, \forall n\in\mathbb{N}.$$

Thus, we need to find $$\lim_{n\to\infty} \frac{1+a_n}{n!}.$$ How to proceed after this?

Answer 1

Let's define $b_n := \dfrac{a_n}{n!}$. Thus, from the recurrence formula:

$$\frac{a_n}{n!}=\frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!}\Rightarrow b_n=b_{n-1}+\frac{1}{(n-1)!}$$

Therefore summing from $1$ to $n-1$, we get:

$$b_n = \sum_{k=0}^{n-1} \frac{1}{k!} \to e$$

Therefore:

$$\lim_{n\to \infty} \frac{a_n+1}{n!}= \lim_{n\to \infty} \frac{a_n}{n!}=\lim_{n\to \infty}b_n = e$$

Answer 2

$$\begin{align} a_{n}&=na_{n-1}+n\\ &=n[(n-1)a_{n-2}+(n-1)]+n\\ &=n(n-1)[(n-2)a_{n-3} + (n-2)]+n(n-1)+n\\ & \phantom{a bit of space here would be nice}\vdots\\ &=\left[n(n-1)(n-2)\dots 3 \cdot 2\cdot 1\right]\left( 1+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{(n-1)!} \right)\\ &=n!\left(1+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{(n-1)!}\right)\rightarrow n!e, \quad (n \to \infty) \end{align} $$