Thread: Proof by Induction

I am to prove the following equation by induction:

(1/2)+(1/6)+(1/12)+...+1/n(n+1)=n/(n+1)

so far I have the base case= 1

Assume this is true for n=k

(1/2)+(1/6)+(1/12)+...+1/k(k+1)=k/(k+1)

now I want to assume this is true for n=k+1

(1/2)+(1/6)+(1/12)+...+1/(k+1)(k+2)=(k+1)/(k+2)

This is where I am stuck...am I placing (k+1) in the correct places?

You have a little problem here:

now I want to assume this is true for n=k+1

(1/2)+(1/6)+(1/12)+...+1/(k+1)(k+2)=(k+1)/(k+2)

You shouldn't assume this. You need to show that this is true, based on the case.

Let's just look at the left hand side and try to work our way to the right.



Look at all of the terms except the last one. By the induction hypothesis, we can substitute them with and receive



Just clean this up and you will get what you want.

How do we get the k/(k+1) on the left hand side of the equation via the induction hypothesis? Do I have to change the right hand side at all or is it just (k+1)/(k+2)?

By the induction hypothesis, you have



As I said earlier, you don't have a right hand side. You just start with
and hope that you can work your way over to .

Originally Posted by baker11108

How do we get the k/(k+1) on the left hand side of the equation via the induction hypothesis? Do I have to change the right hand side at all or is it just (k+1)/(k+2)?

Hi baker11108,

P(k)



P(k+1)



You want to try to prove that the formula being true for n=k causes the formula to be true for n=k+1

Hence we write P(k+1) in terms of P(k) to see if there is a cause.

Therefore if the sum of the first k terms really is

then the sum of k+1 terms is



will be

which is





Since this is what you get when you replace k with k+1 in the RHS
it means that your hypothesis being true for some n=k
causes the hypothesis to be true for the next n=k+1.

Hence "true for n=1 causes it to be true for n=2 causes it to be true for n=3 causes......."

Thanks very much. This helped a lot.