$\displaystyle \displaystyle b_{2010}=\frac{1-b_{2009}}{1+b_{2009}}, \ b_{2009}=\frac{1-b_{2008}}{1+b_{2008}} $
$\displaystyle \displaystyle \frac{1-b_{2009}}{1+b_{2009}}-\frac{1-b_{2008}}{1+b_{2008}}=\frac{(1-b_{2009})(1+b_{2008})-(1-b_{2008})(1+b_{2009})}{(1+b_{2009})(1+b_{2008})}=\ frac{2b_{2008}-2b_{2009}}{(1+b_{2009})(1+b_{2008})}$
The first thing I would do is calculate a few values:
$\displaystyle b_1= \frac{1}{3}$
$\displaystyle b_2= \frac{1- \frac{1}{3}}{1+ \frac{1}{3}}= \frac{2}{3}\frac{3}{4}= \frac{1}{2}$
$\displaystyle b_3= \frac{1- \frac{1}{2}}{1+ \frac{1}{2}}= \frac{1}{2}\frac{2}{3}= \frac{1}{3}$
Of course then
$\displaystyle b_4= \frac{1- \frac{1}{3}}{1+ \frac{1}{3}}= \frac{2}{3}\frac{3}{4}= \frac{1}{2}$
and
$\displaystyle b_5= \frac{1- \frac{1}{2}}{1+ \frac{1}{2}}= \frac{1}{2}\frac{2}{3}= \frac{1}{3}$
Get the point?