Victoria,
There are two steps in a proof by induction. The first is to verify that the statement is true for the smallest value of the index n, in this case it is n = 1.
1
∑ 1/[i(i+1)] = 1/[1(1 + 1)] = 1/2 and n/(n + 1) = 1/(1 + 1) = 1/2
i=1
Hence the statement is true for n = 1.
The second is to show that if the statement is true for any particular value of n then it is also true for the next value of n. To do this I am going to assume that the statement is true for the particular value n = k. That is I assume that
k
∑ 1/[i(i+1)] = k/(k+1) equation 1
i=1
and I want to prove that it is true for the next value of k. Thus I need to show that
k+1
∑ 1/[i(i+1)] = (k+1)/(k+1+1) equation 2
i=1
Here is how I would start this proof.
But from the assumption (equation 1) this can be rewritten
k+1
∑ 1/[i(i+1)] = k/(k+1) + 1/[(k+1)(k+2)]
i=1
Can you simplify this to get (k+1)/(k+2)? If so you will have verified that equation 2 is true and hence, by induction, you have a proof that
n
∑ 1/i(i+1) = n/(n+1)
i=1
is true for all positive integers n.
Penny
