Find $\sum ^{\infty }_{n=2}\dfrac {2+4+6+\ldots +2n}{1+a_{n}}$ Define $a_{n}=1!\left( 1^{1}+1+1\right) +2!\left( 2^{2}+2+1\right) +3!\left( 3^{2}+3+1\right) +\ldots +n!\left( n^{2}+n+1\right) $ , Find $\sum ^{\infty }_{n=2}\dfrac {2+4+6+\ldots +2n}{1+a_{n}}$

You can show that $1+a_n = (n+1) \cdot (n+1)!$ by induction on $n$.
Since $2+4+6+ ... + 2n = n(n+1)$, your sum becomes $$\sum_{n=2}^{\infty} \frac{n}{(n+1)!} = \sum_{n=2}^{\infty} \frac{1}{n!} - \sum_{n=2}^{\infty} \frac{1}{(n+1)!} = \frac{1}{2}.$$
