Yes, it is.
1/(n*(n+1))
= (n+1n)/(n*(n+1))
= (n + 1)/(n*(n+1)) n/(n*(n+1))
= 1/n  1/(n+1)
So, the kth partial  sum has the value
1  1/(k+1)
which's limit for k to infinity is 1 q.e.d.
Sums like these, where a summand is (partly) nihilized by its successor are called telescope sums.
Benjamin
