用户：Jaume Oliver Lafont/四面体数倒数之和

${\displaystyle\sum_{n=1}^{\infty}{\frac{6}{n（n+1）（n+2）}}=\sum_{n=1}^{\fraty}{\frac{3}{n$

${\displaystyle=\sum_{n=1}^{\infty}{\frac{3}{n（n+1）}}-\sum_{n=2}^{\frity}{\frac{3}}$

$｛\displaystyle＝｛\frac｛3｝｛2｝｝+\sum_｛n=2｝^｛\frac｛3｝｛n（n+1）｝-\sum_｛n=2｝^｛\frac｝｛3｝｛n（n+1）｝＝｛\frac｛3｝｛2｝｝｝$

(A000292号)