Yes, it is.
1/(n*(n+1))
= (n+1-n)/(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
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