Re: Sigma Proof by Induction

Quote:

Originally Posted by

**renolovexoxo** Prove 1/1(2)+2/2(3)+...+1/n(n+1)=n/n+1

I've been using induction to prove this.

For the base case I got: 1/1(2)=1/2

Then I go on and get a little more stuck:

Assume 1/1(2)+2/2(3)+...+1/n(n+1)=n/n+1 for all n in N. Want to show that 1/1(2)+2/2(3)+...+1/n(n+1)+1/(n+1)(n+2)=(n+1)/(n+1)+1

I simplified it using the assumption

n/(n+1)+1/(n+1)(n+2)=(n+1)/(n+1)+1

Then when I tried to simplify I keep getting something that isn't an equality. Have I done something wrong so far?

Notice that , so your sum

Re: Sigma Proof by Induction

Quote:

Originally Posted by

**Prove It** Notice that

, so your sum

I have to prove by induction though, so I couldn't use this approach, correct?

Re: Sigma Proof by Induction

Quote:

Originally Posted by

**renolovexoxo** I have to prove by induction though, so I couldn't use this approach, correct?

Here is what is needed.

Re: Sigma Proof by Induction

Quote:

Originally Posted by

**renolovexoxo** I have to prove by induction though, so I couldn't use this approach, correct?

That's not what you said. You said that you were trying to prove the statement and that you used an induction approach. I just showed you an easier approach...

Re: Sigma Proof by Induction

I'm sorry, then I misstated my intentions. I need to use induction to complete this proof.

Re: Sigma Proof by Induction

Quote:

Originally Posted by

**renolovexoxo** I'm sorry, then I misstated my intentions. I need to use induction to complete this proof.

Did you follow reply #4?

Re: Sigma Proof by Induction

I understood it, if that is what you're asking.