1. ## Inequality

Can somebody demonstrate:
$\displaystyle \ \frac{n}{{\sqrt[n]{{n!}}}} < \left( {1 + \frac{1}{n}} \right)^n \$

2. First we write the inequality in the equivalent form: $\displaystyle \log \left( n \right) - n \cdot \log \left( {1 + \tfrac{1} {n}} \right) < \tfrac{1} {n} \cdot \log \left( {n!} \right)$

Note that $\displaystyle \log \left( {1 + \tfrac{1} {n}} \right) \geq \tfrac{1} {n} - \tfrac{1} {2} \cdot \left( {\tfrac{1} {n}} \right)^2$ - think about the Taylor series, or you can also prove that $\displaystyle \log(1+x)\geq{x-\tfrac{x^2}{2}}$ (x>0) by differentiating -

Hence we have: $\displaystyle LHS= \log \left( n \right) - n \cdot \log \left( {1 + \tfrac{1} {n}} \right) \leq \log \left( n \right) - 1 + \tfrac{1} {{2n}}$

Next note that: $\displaystyle \log \left( {n!} \right) = \sum\limits_{k = 1}^n {\log \left( k \right)} = \int_0^n {\log \left( x \right)dx} + \int_0^n {\tfrac{{\left\{ x \right\}}} {x}dx}$ - see Lemma 1, here-

So, since $\displaystyle \int_0^n {\tfrac{{\left\{ x \right\}}} {x}dx} \geqslant \int_0^1 {\tfrac{{\left\{ x \right\}}} {x}dx} = \int_0^1 {dx} = 1$ we find: $\displaystyle \log \left( {n!} \right) \geq n \cdot \log \left( n \right)-n + 1$

Hence: $\displaystyle \log \left( n \right) - 1 + \tfrac{1} {n} \leq \tfrac{1} {n} \cdot \log \left( {n!} \right) = RHS$ and finally : $\displaystyle \log \left( n \right) - n \cdot \log \left( {1 + \tfrac{1} {n}} \right) \leq \log \left( n \right) - 1 + \tfrac{1} {{2n}} < \log \left( n \right) - 1 + \tfrac{1} {n} \leq \tfrac{1} {n} \cdot \log \left( {n!} \right)$

In fact, the LHS of your inequality ( as written originally) converges to 'e'.

3. It's seem be ok,only i have one question:
WHY
$\displaystyle \int_0^n {\tfrac{{\left\{ x \right\}}} {x}dx} \geqslant \int_0^1 {\tfrac{{\left\{ x \right\}}} {x}dx} = \int_0^1 {dx}$

4. Because $\displaystyle \tfrac{{\left\{ x \right\}}} {x}\geq 0$ when $\displaystyle x\in (0,+\infty)$ implies that the sequence $\displaystyle a_n=\int_0^n {\tfrac{{\left\{ x \right\}}} {x}dx}$ is non-decreasing ( $\displaystyle n\in {\mathbb{Z}^+}$ ) and so $\displaystyle a_n\geq{a_1}$.

Now when $\displaystyle x\in (0, 1)$ we have $\displaystyle {\left\{ x \right\}}=x$, and so $\displaystyle \int_0^1 {\tfrac{{\left\{ x \right\}}} {x}dx} = \int_0^1 {dx}=1$