How we can prove that is monotone and bounded: $I_n=\int _1^n\:e^{-x^3}dx\:$ , Have any ideea how we can solve? and explain all to understand, I am a student...
P.S: for all guys on this site, you do a great job and I respect all because we help each other, and I don't want to see somebody who don't help but makes clever with me... I am at the beginning and don't know...
To show that the sequence is monotone, consider the difference of two consecutive terms: $$ I_{n+1} - I_n = \int_1^{n+1} - \int_1^n = \int_n^{n+1} e^{-x^3} dx > 0 $$ which shows that $I_{n+1} > I_n$ for all $n$. We don't need to perform any integration to show this.
To show that the sequence is bounded, we must find a constant such that all $I_n$ are less or equal than this number. To do that, an integration will be necessary, but since we only have to find a (crude) estimate, we are free to change the integrand to make the integral easier, as long as an upper bound results in the end.
Note that $x^3 \ge x$ for $x \ge 1$ and therefore $e^{-x^3} \le e^{-x}$ for $x \ge 1$. Hence for any $n > 1$ $$ \int_1^{n} e^{-x^3}dx \le \int_1^n e^{-x} dx = e^{-1} - e^{-n} < e^{-1} . $$ So all $I_n$ are bounded from above by the same constant.