prove by induction that 1/1(2)+1/2(3)+1/3(4)+... + 1/n(n+1)=n/(n+1)

Question

1. show case n=1 is true

2. assume case=k is true

3. prove case k+1 is true

Answer 1

1. Base case: n=1, S₁=1/(1×2)=1/2=1/(1+1). So base case is true.
2. If a[k]=1/(k(k+1)) represents the kth term, and we assume the sum to the kth term is S[k]=k/(k+1), then S[k+1]=S[k]+a[k+1]=k/(k+1)+1/((k+1)(k+2))=(k(k+2)+1)/((k+1)(k+2)).
3. So we get (k²+2k+1)/((k+1)(k+2)) = (k+1)²/((k+1)(k+2)) = (k+1)/(k+2)=S[k+1]. This proves by induction that S[n]=n/(n+1).