Proof:
(1) Since $$2a_{n+1}-a_{n}= \frac {n-2}{n(n+1)(n+2)},$$ $$b_{n}=a_{n}- \frac {1}{n(n+1)},$$
we have $a_{n+1}= \frac {1}{2}a_{n}+ \frac {n-2}{2n(n+1)(n+2)}.$
Therefore, $$\frac {b_{n+1}}{b_{n}}= \frac { \frac {1}{2}a_{n}+ \frac {n-2}{2n(n+1)(n+2)}- \frac {1}{(n+1)(n+2)}}{a_{n}- \frac {1}{n(n+1)}}= \frac { \frac {1}{2}a_{n}- \frac {1}{2}× \frac {1}{n(n+1)}}{a_{n}- \frac {1}{n(n+1)}}= \frac {1}{2}.$$
Therefore, $\{b_n\}$ is a geometric sequence with a common ratio of $\frac {1}{2}$, and the first term is $a_1- \frac {1}{2}= \frac {1}{2}.$
Thus, $b_n= \left( \frac {1}{2} \right)^{n}.$
Hence, $a_n= \left( \frac {1}{2} \right)^{n}+ \frac {1}{n(n+1)}.$

(2) $C_n=n b_n=n \times \left( \frac {1}{2} \right)^{n}.$
Therefore, $T_n= \frac {1}{2}+2 \times \frac {1}{2^{2}}+3 \times \frac {1}{2^{3}}+\ldots+n \times \frac {1}{2^{n}},$
$\frac {1}{2}T_n= \frac {1}{2^{2}}+2 \times \frac {1}{2^{3}}+\ldots+(n-1) \times \frac {1}{2^{n}}+n \times \frac {1}{2^{n+1}}.$
Thus, $\frac {1}{2}T_n= \frac {1}{2}+ \left( \frac {1}{2} \right)^{2}+\ldots+\left( \frac {1}{2} \right)^{n}-n \times \frac {1}{2^{n+1}}= \frac { \frac {1}{2}(1- \frac {1}{2^{n}})}{1- \frac {1}{2}}-n \times \frac {1}{2^{n+1}}=1- \frac {2+n}{2^{n+1}}.$
Therefore, $T_n=2- \frac {2+n}{2^{n}}<2.$
Thus, we conclude that $T_n<2$ with $\boxed{T_n<2}$.