$$\because\quad\frac{2}{n^2 + 4n +3}=\frac{2}{(n+1)(n+3)}=\frac{1}{n+1}-\frac{1}{n+3}$$ $$\therefore\quad\sum_{n=1}^{\infty}\frac{2}{n^2 + 4n +3}=\frac{1}{1+1}+\frac{1}{2+1}=\boxed{\frac{5}{6}}$$