Proof by Induction: 1/1*2 + 1/2*3 + 1/3*4 +...+ 1/n(n+1) = n/n+1

[mweaver68]

trying to prove the following

1/1*2 + 1/2*3 + 1/3*4 +...+ 1/n(n+1) = n/n+1

Prove P(1) true: 1/1*2 = 1/1+1 = 1/2

Assume P(k) true: 1/2 + 1/6 + 1/12 + ... + 1/k(k+1) = k/(k+1)

trying to prove P(k+1) true:

step 1: 1/2 + 1/6 + 1/12 + ... + 1/k(k+1) + 1/(k+1)[(K+1)+1] = k+1/[(k+1) + 1]
step 2: k / (k+1) + 1/(k+1)[(K+1)+1] = k+1/[(k+1) + 1]

I keep ending up with k+1 / (k+2) ^2 on the left hand side.

I have not been able to figure out what I am doing wrong here.

Any ideas.

Thanks.

 

[HallsofIvy]

trying to prove the following

1/1*2 + 1/2*3 + 1/3*4 +...+ 1/n(n+1) = n/n+1

Prove P(1) true: 1/1*2 = 1/1+1 = 1/2

Assume P(k) true: 1/2 + 1/6 + 1/12 + ... + 1/k(k+1) = k/(k+1)

trying to prove P(k+1) true:

step 1: 1/2 + 1/6 + 1/12 + ... + 1/k(k+1) + 1/(k+1)[(K+1)+1] = k+1/[(k+1) + 1]

You don't know this yet, this is what you are trying to prove. You know that
1/2+ 1/6+ 1/12+ ...+ 1/[k(k+1)]+ 1/[(k+1)((k+1)+1)]= k/(k+1)+ 1/[(k+1)(k+2).
Now the right hand side WILL have k²+ 2k+ 1= (k+1)² in the numerator, but that's easily taken care of!

step 2: k / (k+1) + 1/(k+1)[(K+1)+1] = k+1/[(k+1) + 1]

I keep ending up with k+1 / (k+2) ^2 on the left hand side.

I have not been able to figure out what I am doing wrong here.

Any ideas.

Thanks.

 

[mweaver68]

Thanks for the reply.

You stated "1/2+ 1/6+ 1/12+ ...+ 1/[k(k+1)]+ 1/[(k+1)((k+1)+1)]= k/(k+1)+ 1/[(k+1)(k+2)." but I am not sure how you got the right side of this. I thought when trying to prove P(k+1), you substitute k+1 for all k's on the right side. so why isn't it (k+1) / [(k+1) + 1]?