Evaluating $\int\limits_{1}^{\infty} \frac{1}{\lfloor{x}\rfloor!}dx$

Question

I thought the improper integral $\int\limits_1^{\infty} \frac{1}{\lfloor{x}\rfloor!}dx$ converge, while the textbook says it's not.
Of course, $\lfloor{x}\rfloor$ is the greatest integer function. Here is my solution:

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For any $n\in \mathbb N$,$$\lfloor{x}\rfloor=n\;\; \Leftrightarrow \;\; n\leq x<n+1 $$
Thus if $\; N\leq t<N+1$ $$\begin{align}&\int_{1}^{N}\frac{1}{\lfloor{x}\rfloor!}dx\leq\int_{1}^{t}\frac{1}{\lfloor{x}\rfloor!}dx\leq\int_{1}^{N+1}\frac{1}{\lfloor{x}\rfloor!}dx\\ &\Rightarrow \;\sum_{n=1}^{N-1}\frac{1}{n!}\leq\int_{1}^{t}\frac{1}{\lfloor{x}\rfloor!}dx \leq\sum_{n=1}^{N}\frac{1}{n!}\end{align}$$ Taking $N\rightarrow\infty$ both sides also gives $t\rightarrow\infty$, and we have $$e-1\leq \int_1^{\infty} \frac{1}{\lfloor{x}\rfloor!}dx \leq e-1$$ Therefore, $\int_1^{\infty} \frac{1}{\lfloor{x}\rfloor!}dx=e-1$. □

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I've tried a sort of times to find some mistakes in what I wrote. But I didn't get anything till now.
Can somebody point out what I missed? Thank you.

Answer 1

Setting $$ I=[1,\infty), \quad I_n=[n,n+1) \quad \forall n\ge1, $$ we have $$ I=\bigcup_{n=1}^{\infty}I_n, \quad I_j\cap I_k=\varnothing \quad \forall j\ne k. $$ Since $$ \lfloor x \rfloor =n \quad \forall x \in I_n, \quad n=1,2,3, \ldots $$ it follows that \begin{eqnarray} \int_1^{\infty}\frac{dx}{\lfloor x\rfloor !} &=&\int_I\frac{dx}{\lfloor x \rfloor ! }\cr &=&\sum_{n=1}^{\infty}\int_{I_n}\frac{dx}{\lfloor x \rfloor !}\cr &=&\sum_{n=1}^{\infty}\int_{I_n}\frac{dx}{n!}\cr &=&\sum_{n=1}^{\infty}\frac{n+1-n}{n!}\cr &=&\sum_{n=1}^{\infty}\frac{1}{n!}\cr &=&-1+\sum_{n=0}^{\infty}\frac{1}{n!}\cr &=& e-1. \end{eqnarray}

Answer 2

$$ \begin{aligned} \int_{1}^{∞} \frac{1}{\left \lfloor x \right \rfloor !} dx&=\lim_{N \to ∞}\lim_{λ_{max} \to 0} \sum_{i=1}^{N} \frac{λ_{i}}{\left \lfloor x_{i} \right \rfloor !} \\ &=\lim_{N \to ∞} \sum_{n=1}^{N} \frac{1}{n!} \\ &=e-1. \end{aligned} $$