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For a given sequence, $a_1=1$ and $a_n=n(1+a_{n-1})$ $\forall n\geq 2$, then value of given limit:

$$\lim_{n\to \infty} \bigg(1+\frac{1}{a_1}\bigg)\bigg(1+\frac{1}{a_2}\bigg)\cdots\bigg(1+\frac{1}{a_n}\bigg)$$

Usually such type of questions are solved by squeeze theorem or by converting them into definite integral but don't see neither working here. Could someone give me little help to proceed

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Since $a_1=1$ and $$ 1+\frac1{a_{n-1}}=\frac{a_n}{na_{n-1}}\tag1 $$ induction says $$ \prod_{k=1}^{n-1}\left(1+\frac1{a_k}\right)=\frac{a_n}{n!}\tag2 $$ Furthermore, dividing the recursion by $n!$ yields $$ \frac{a_n}{n!}=\frac{a_{n-1}}{(n-1)!}+\frac1{(n-1)!}\tag3 $$ which, by induction, gives $$ \frac{a_n}{n!}=\sum_{k=0}^{n-1}\frac1{k!}\tag4 $$ Combining $(2)$ and $(4)$, we have $$ \prod_{k=1}^\infty\left(1+\frac1{a_k}\right)=\sum_{k=0}^\infty\frac1{k!}\tag5 $$

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Hint: $\displaystyle \frac{a_n}{n!} = \frac{a_{n - 1}}{(n - 1)!} + \frac{1}{(n - 1)!} \ (n \geqslant 2)$ and$$ \prod_{k = 1}^n \left(1 + \frac{1}{a_k}\right) = \left. \prod_{k = 1}^n (1 + a_k) \middle/ \prod_{k = 1}^n a_k\right. = \left.\prod_{k = 1}^n \frac{a_{k + 1}}{k + 1} \middle/ \prod_{k = 1}^n a_k\right. = \frac{1}{a_1} \frac{a_{n + 1}}{(n + 1)!}. $$

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