Proving the Summation Formula for 1/(n(n+1)) Using Mathematical Induction

[knowledgeSeeker]

Need help with proof by mathematical induction that (1/(1*2)) + (1/(2*3)) + ... + (1/(n(n+1)) = (n/(n+ 1)) for all integers n >= 1.

Basis step: for n = 1: (1/(1*2)) = 1/2 and (1/(1+1) = 1/2, hence property is true for n = 1.

Inductive step: want to show that for alll integers k >= 1, if n = k is true then n = k + 1 is true. How do I prove? Believe I want to show (1/(1*2)) + (1/(2*3)) + [1/((k+1)((k+1)+1)] = [(k + 1)/((k+1) + 1)], but how??

Thank you for any suggestions.

 

[gnome]

So you have shown that P(1) is true. Now you want to show that if you assume that P(k) is true, it follows that P(k+1) is true. So first write the expression for P(k), which you assume to be true. Then add the next number in the series (to both sides), and see if you can rearrange the expression on the right side into the form that you are trying to prove.

 

[HallsofIvy]

Let Sk= 1/(1*2)+ 1/(2*3)+ ...+ 1/(k(k+1)), the sum for n= k
Then S(k+1)= 1/(1*2)+ ...+ 1/(k)(k+1)+ 1/((k+1)((k+1)+1)= Sk+ 1/((k+1)(k+2))

By your "induction hypothesis", Sk= k/(k+1).

What is k/(k+1)+ 1/((k+1)(k+2)) ?

 

[knowledgeSeeker]

Thank you. Proved both sides = (k+1)/(k+2). Hence, true for n = k +1 and since both basis and inductive steps true, true for all n >= 1.