Question 1065476
Look at a partial sum from 1 to k.
{{{sum(2/(n(n+1)(n+2)),n=1,k)=(k(k+3))/(2(k+1)(k+2))}}}
{{{sum(2/(n(n+1)(n+2)),n=1,k)=(k^2+3k)/(2(k^2+3k+2))}}}

If we take the limit as k approaches infinity,
{{{lim(k->infinity,(k^2+3k)/(2(k^2+3k+2)))=lim(k->infinity,(1+3/k)/(2(1+3/k+k^2)))}}}
{{{lim(k->infinity,(k^2+3k)/(2(k^2+3k+2)))=(1+0)/(2(1+0+0)))}}}
{{{lim(k->infinity,(k^2+3k)/(2(k^2+3k+2)))=highlight(1/2)}}}