Limit $(1+\frac{1}{a_n})(1+\frac{1}{a_{n-1}} )\cdots(1+\frac{1}{a_1}) $ where $a_n=n(1+a_{n-1})$ and $a_1 =1$ Suppose $a_n=n(1+a_{n-1})$ and $a_1 =1$. Then the limit of $(1+\frac{1}{a_n})(1+\frac{1}{a_{n-1}} )\cdots(1+\frac{1}{a_1}) $ where $n$ tends to infinity is? I got $1/2$ for answer is it correct?
 A: $$
\begin{align}
&\left(1+\frac1{a_1}\right)\left(1+\frac1{a_2}\right)\left(1+\frac1{a_3}\right)\cdots\left(1+\frac1{a_{n-1}}\right)\\
&=\frac{a_1+1}{a_1}\frac{a_2+1}{a_2}\frac{a_3+1}{a_3}\cdots\frac{a_{n-1}+1}{a_{n-1}}\\
&=\frac{a_2/2}{a_1}\frac{a_3/3}{a_2}\frac{a_4/4}{a_3}\cdots\frac{a_n/n}{a_{n-1}}\\
&=\frac{a_n}{a_1}\frac1{n!}\tag{1}
\end{align}
$$
Induction shows that
$$
a_n=n!\sum_{k=0}^{n-1}\frac1{k!}\tag{2}
$$
Plug $(2)$ into $(1)$ and we get 
$$
\begin{align}
&\lim_{n\to\infty}\left(1+\frac1{a_1}\right)\left(1+\frac1{a_2}\right)\left(1+\frac1{a_3}\right)\cdots\left(1+\frac1{a_{n-1}}\right)\\
&=\lim_{n\to\infty}\frac{a_n}{a_1}\frac1{n!}\\
&=\sum_{k=0}^\infty\frac1{k!}\\[6pt]
&=e\tag{3}
\end{align}
$$
A: EDIT: Sorry, ignore my previous post.  I realized something that makes my previous answer invalid.
I think the answer must be greater than 1.  The first term,
$$\left( 1 + \frac{1}{a_1} \right) = 1 + \frac{1}{1} = 1 + 1 = 2 $$
Moreover, $a_n$ is a strictly positive sequence, so for all $n$,
$$ 1 + \frac{1}{a_n} > 1 $$
If your first term in the infinite product is 2, there is no way you can reduce the product by multiplying on terms that are greater than 1.
