The above formula =1[(n+2) c ((n+1), 2)] = 2/[(n+2) (n+1) n] =1[n (n+)
Divide each item as above, then add them together. Note that the last formula of each item can offset the previous formula of the next item, so
An =1(1x 2)-1/[(n+1) (n+2)], when n is finite, an= 1/2.