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


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, $\int_1^{\infty} \frac{1}{\lfloor{x}\rfloor!}dx=e-1$. □


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.

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    $\begingroup$ Yes, it converges. Are you sure your textbook says that it does not ? $\endgroup$ – TheSilverDoe Feb 23 at 9:21
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    $\begingroup$ Your work is very correct $\to +1$ $\endgroup$ – Claude Leibovici Feb 23 at 9:33
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    $\begingroup$ So: either a mistake in the the textbook, or the OP miscopied the problem here. Since the identity of the textbook is secret, no one here can verify this. $\endgroup$ – GEdgar Feb 23 at 12:54
  • $\begingroup$ Thank you guys all! Later I will call the author to verify it. $\endgroup$ – JJLEE Feb 23 at 15:42

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