Compare the series to the convergent series,
[tex]\displaystyle\sum_{n=1}^\infty\frac1{n^2}[/tex]
By the limit comparison test, the given series converges because
[tex]\displaystyle\lim_{n\to\infty}\frac{\frac{n+1}{n(n+2)(n+3)}}{\frac1{n^2}}=\lim_{n\to\infty}\frac{n(n+1)}{(n+2)(n+3)}=1[/tex]
