show that the sequence {$b_n$}, where $b_n = ( 1 + \frac {x}{n})^{l+n}$ for $n \in \mathbb{N}$, is strictly decreasing. suppose that  $x > 0$, $l \in \mathbb{N}$ and $l > x.$ show that the  sequence {$b_n$}, where  $b_n = ( 1 + \frac {x}{n})^{l+n}$  for  $n \in \mathbb{N}$,  is  strictly  decreasing.
My attempts :   i take $b_{n+1} - b_n =  ( 1 + \frac {x}{n+1})^{l+n+1}-   ( 1 + \frac {x}{n})^{l+n}$......as  i don't know  how to proceed  Further
Pliz help me   any Hinst/solution will be appreciated,,
thanks u 

$l>x\,$ , $\,n>0\,$ , $\,\displaystyle (1+\frac{x}{n})^n<e^x\,$ :
$\displaystyle \frac{d}{dn}\ln b_n = \frac{1}{n}\left(\ln \left((1+\frac{x}{n})^n\right)-\frac{x(l+n)}{n+x}\right) < 
\frac{x}{n}\left(1-\frac{n+l}{n+x}\right)<0$ 
$ \ln b_n $ is strictly decreasing therefore $\, b_n>0\,$ is strictly decreasing too: 
$\displaystyle \frac{d}{dn}b_n = b_n \frac{d}{dn}\ln b_n<0$ 
