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Evaluate $L=\displaystyle\lim_{n \rightarrow \infty} \left(1+\dfrac{1}{a_1}\right) \left(1+\dfrac{1}{a_2}\right) \dots \dots \left(1+\dfrac{1}{a_n }\right)$ where $a_1=1$ and $a_n=n(1+a_{n-1}), \ \forall \ n \geq 2$

By rewriting $L=\displaystyle\lim_{n \rightarrow \infty} \left(\dfrac{1+a_1}{a_1}\right) \left(\dfrac{1+a_2}{a_2}\right) \dots \dots \left(\dfrac{1+a_n}{a_n }\right)$ and observing that $1+a_m=\dfrac{a_{m+1}}{m+1}$, I reduced $L$ to $$L=\displaystyle\lim_{n \rightarrow \infty}\dfrac{a_{n+1}}{(n+1)!}$$

I do not know know how to proceed further.

Solutions and hints in the right direction would be appreciated.

user1001001
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2 Answers2

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$$\frac{a_{n+1}}{(n+1)!}=\frac{1}{n!}+\frac{a_n}{n!}= \frac{1}{n!}+\frac{1}{(n-1)!}+\frac{a_{n-1}}{(n-1)!}= \frac{1}{n!}+\frac{1}{(n-1)!}+ \cdots +\frac{1}{2!}+1+1 $$

So $$\lim \frac{a_{n+1}}{(n+1)!}=e$$

Rene Schipperus
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\begin{align} \lim_{n \to \infty}\dfrac{a_{n+1}}{(n+1)!}&= \lim_{n \to\infty}\dfrac{{(n+1)(1+a_n)}}{(n+1)!}\\ &= \lim_{n \to\infty}(\dfrac{a_n}{n!}+\dfrac{1}{n!})\\ &= \lim_{n \to\infty}(\dfrac{a_{n-1}}{(n-1)!}+\dfrac{1}{(n-1)!}+\dfrac{1}{n!})\\ &\vdots\\&= \lim_{n \to\infty}(1+\dfrac{1}{{1}!}+\dfrac{1}{{2}!}+\ldots+\dfrac{1}{n!})\\ &=e \end{align}

user62498
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