# If $S_{n}=\large{1}{6}\;n(n+1)(n+2)$ $\forall\;n\;\geq\;1$, then $lim_{n \to \infty}\;\sum_{r=1}^{n}\;\large\frac{1}{a_{r}}$ equals

$(a)\;1\qquad(b)\;2\qquad(c)\;3\qquad(d)\;6$

Explanation : $a_{n}=S_{n}-S_{n-1}$
$=\large\frac{1}{6}\;n(n+1)(n+2)-\large\frac{1}{6}\;(n-1)(n+1)n$
$=\large\frac{1}{6}\;n(n+1)(n+2-n+1)$
$=\large\frac{1}{2}\;n(n+1)$
$\sum_{r=1}^{n}\;\large\frac{1}{a_{r}}=\sum_{r=1}^{n}\large\frac{2}{r(r+1)}$
$\lim_{n \to \infty}\;\sum_{r=1}^{n}\;\large\frac{2}{r(r+1)}$
$=\sum_{r=1}^{\infty}\;2\;[\large\frac{1}{r}-\large\frac{1}{r+1}]$
$=\large\frac{2}{1}=2.$