Q:

Calculating limit of a particular product series

How to find 
$$\lim\limits_{n\to\infty} \left(1+\frac{ 1 }{ a_{ 1 } }  \right) \left( 1+\frac { 1 }{ a_{ 2} }  \right)\cdots\left( 1+\frac { 1 }{ a_{ n } }  \right) $$ 
where 
$$a_1=1$$ 
$$a_n=n(1+a_{n-1})$$ 
for all $n \geq 2$?

A:

When looking for the limit of an infinite product (or a series), it is often helpful to look for cancellation/telescoping. If telescoping occurs, that can give you a much simpler expression for the partial products (partial sums), from which the limit may be easy to determine.
Here, rewriting the factors as
$$\frac{a_k + 1}{a_k}$$
and using the recurrence $a_k = k(1+a_{k-1})$ (for $k > 1$) gives us
$$\prod_{k=1}^n \biggl(1 + \frac{1}{a_k}\biggr) = \frac{1+a_1}{a_1}\prod_{k = 2}^n \frac{a_k+1}{k(a_{k-1}+1)} = \frac{a_n + 1}{n!}.$$
So we see that we want to find $\lim\limits_{n\to\infty} \dfrac{a_n}{n!}$. For those terms, the recurrence becomes
$$\frac{a_n}{n!} = \frac{1}{(n-1)!} + \frac{a_{n-1}}{(n-1)!},$$
which together with the initial value $\frac{a_1}{1!} = \frac{1}{0!}$
 yields
$$\frac{a_n}{n!} = \sum_{k = 0}^{n-1} \frac{1}{k!},$$
from which the limit $e$ is immediately read off.

