Evaluate the following infinite product: $$\lim_{n\to \infty} \prod_{r=1}^n (1+ \frac{1}{a_r}),\\ a_1=1; \quad a_r=r(1+a_{r-1}).$$
I thought of two ways in which this might be evaluated: (1) By taking log and converting into a Riemann series which I saw isn't possible and (2) by converting into a polynomial equation of nth degree where coefficients become sums as per the theory of equations. Right now, I'm stuck, trying to figure out the way to proceed.
As @Gribouillis mentionned, it is generally a good reflex to try to fit the recurrence equation in the product (or the sum). Here you have $$\prod_{k=1}^n (1+\frac{1}{a_k}) = \prod_{k=1}^n \frac{1+a_k}{a_k} = \prod_{k=1}^n \frac{a_{k+1}}{(k+1)a_k} = \frac{a_{n+1}}{(n+1)!a_1}$$ Let's see if we can find a limit for this. Note $b_n=\frac{a_n}{n!}$. With the recurrence relation, you get, by dividing by $n!$ : $$b_n=\frac{a_n}{n!} = \frac{1}{(n-1)!}(1+a_{n-1}) = \frac{1}{(n-1)!} + \frac{a_{n-1}}{(n-1)!} = \frac{1}{(n-1)!} + b_{n-1}$$ So $$b_n=b_1+\sum_{k=2}^n\frac{1}{(k-1)!} = \sum_{k=0}^{n-1}\frac{1}{(k-1)!}$$ and it is well known that $(b_n)$ has limit $e$.