1. ## Inequality

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

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

Note that $
\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 $\log(1+x)\geq{x-\tfrac{x^2}{2}}$ (x>0) by differentiating -

Hence we have: $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: $
\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 $
\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: $
\log \left( {n!} \right) \geq n \cdot \log \left( n \right)-n + 1
$

Hence: $
\log \left( n \right) - 1 + \tfrac{1}
{n} \leq \tfrac{1}
{n} \cdot \log \left( {n!} \right) = RHS
$
and finally : $
\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
$
\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 $\tfrac{{\left\{ x \right\}}} {x}\geq 0$ when $x\in (0,+\infty)$ implies that the sequence $a_n=\int_0^n {\tfrac{{\left\{ x \right\}}} {x}dx}$ is non-decreasing ( $n\in {\mathbb{Z}^+}$ ) and so $a_n\geq{a_1}$.

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