Verify the identity 1/(k*(k+1)) = 1/k - 1/(k+1) and use Exercise 11 (telescoping series) to compute the sum from k=1 to n of 1/(k*(k+1)).

Telescoping series or sum means the sum of a sequence of terms of the form a_j - a_(j-1). The idea is simple: if you expand it you get cancellation.
(a_1-a_0) + (a_2-a_1) + (a_3-a_2) + ... + (a_n - a_(n-1) )
simplifies to just a_n-a_0 because if you just rearrange all the terms and do all the obvious cancelling, this is all that's left.

So we can sum 1/(k*(k+1)) for k from 1 to n, by realizing that this is the same as (a_1-a_0) + (a_2-a_1) + (a_3-a_2) + ... + (a_n - a_(n-1) )
where we let a_j= -(1/j). Why the minus sign? Well, our sum is of the terms
1/(j*(j+1)) = 1/j - 1/(j+1) = (-1/(j+1) - (-1/j) = a_j - a_(j+1) .
It is probably simpler to write this out directly:

1/1*2 + 1/2*3 + ... + 1/n*(n+1) = (1/1-1/2) + (1/2-1/3) + ... + (1/n - 1/(n+1))
= (1/1) - (1/(n+1)) It remains to verify that 1/(k*(k+1)) is equal to 1/k - 1/(k+1). Just add the fractions together by putting them over the common denominator k*(k+1).