A calculus problem by Anish Roy

We define another sequence { $b_{n}$ } as $b_{n}=\frac{a_{n}}{n!}$ . Then note that for $n\geq2$ , $b_{n}=\frac{a_{n}}{n!}=\frac{a_{n-1}+1}{(n-1)!}=b_{n-1}+\frac{1}{(n-1)!} \Rightarrow b_{n}-b_{n-1}=\frac{1}{(n-1)!}$

Therefore $b_{n}=b_{1}+\displaystyle \sum_{k=2}^{n}(b_{k}-b_{k-1})=1+\displaystyle \sum_{k=2}^{n}\frac{1}{(k-1)!}=\displaystyle \sum_{k=0}^{n-1}\frac{1}{k!}$ Hence $\begin{aligned} \displaystyle \lim_{n \to \infty}P_{n}= & \displaystyle \lim_{n \to \infty}(1+\frac{1}{a_{1}})(1+\frac{1}{a_{2}})\cdots(1+\frac{1}{a_{n}}) \\ & =\displaystyle \lim_{n \to \infty}(\frac{a_{1}+1}{a_{1}}.\frac{a_{2}+1}{a_{2}}\cdots \frac{a_{n}+1}{a_{n}}) \\ & =\displaystyle \lim_{n \to \infty}(\frac{a_{2}}{2a_{1}}.\frac{a_{3}}{3a_{2}}\cdots \frac{a_{n+1}}{(n+1)a_{n}}) \\ & =\displaystyle \lim_{n \to \infty} \frac{a_{n+1}}{(n+1)!} \\ & =\displaystyle \sum_{k=0}^{\infty}\frac{1}{n!} = e \end{aligned}$ $\therefore x=e=2.71828$

Thus $\lfloor{x}\rfloor=\lfloor{e}\rfloor=\lfloor{2.71828}\rfloor=\boxed{2}$