How to prove that $1 - \frac{1}{n} < \int\limits_0^1 e^{-x^n}dx < 1, n > 1$ How to prove that $1 - \frac{1}{n} < \int\limits_0^1 e^{-x^n}dx < 1, n > 1$
How prove tasks like this?
 A: We have, forall $0 <x\leq 1$ and $n>1$,
by MVT,
$$e^{-x^n}-1=-x^ne^{-c} $$ with $$0 <c <x^n \leq 1$$
then
$0 <e^{-c}<1$
and
$-x^n<e^{-x^n}-1 <0$
thus
$$\int_0^1-x^ndx <\int_0^1 (e^{-x^n}-1)dx<0$$
and
$$-\frac {1}{n}<-\frac {1}{n+1}<I-1 <0$$
where $I $ is your integral.
A: For any $n>1$ we have $e^{-x^n}=1-x^n+O(x^{2n})$ for all $x\in[0,1]$, then we have 
$$
1-\frac{1}{n} < 1-\frac{1}{n+1}=\int_0^1 1-x^n\,dx \le\int_0^1e^{-x^n}\,dx < \int_0^1 1\,dx = 1.
$$
A: Well, we know that:
$$e^x=\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\tag1$$
So, we also get:
$$e^{-x^\text{n}}=\sum_{\text{k}=0}^\infty\frac{\left(-x^\text{n}\right)^\text{k}}{\text{k}!}=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot x^{\text{n}\text{k}}}{\text{k}!}\tag2$$
So, when we look at the integral:
$$\int_0^1e^{-x^\text{n}}\space\text{d}x=\int_0^1\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot x^{\text{n}\text{k}}}{\text{k}!}\space\text{d}x=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\int_0^1x^{\text{n}\text{k}}\space\text{d}x=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\left(\frac{1^{1+\text{n}\text{k}}}{1+\text{n}\text{k}}-\frac{0^{1+\text{n}\text{k}}}{1+\text{n}\text{k}}\right)=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{1}{1+\text{n}\text{k}}=\frac{\Gamma\left(\frac{1}{\text{n}},0,1\right)}{\text{n}}\tag3$$
A: $e^{-x}$, as well as $e^{-x^n}$ for $n\in\mathbb{N}^+$, is an entire function. It follows that
$$ I_n = \int_{0}^{1} e^{-x^n}\,dx = \int_{0}^{1}\sum_{m\geq 0}\frac{(-1)^m x^{mn}}{m!}\,dx = 1-\sum_{m\geq 1}\frac{(-1)^{m+1}}{m!(mn+1)} $$
and since $g(m)=\frac{1}{m!(mn+1)}$ is a rapidly decreasing function on $\mathbb{N}^+$ we have:
$$ 1-\frac{1}{n+1}\leq I_n \leq 1-\frac{1}{n+1}+\frac{1}{4n+2}.$$
