Inductive proof of summation formula

[livcon]

Homework Statement

Prove by induction the following summation formula:
\frac{1}{1\1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}
n \geq 1

Homework Equations

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The Attempt at a Solution

Inductive step:
1. \frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{(n+1)+1}

2. \frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}

To prove this (1) should equal (2) (right?), but I don't see how to manipulate (2) to make it equal to (1)

Thanks in advance!

 

[whs]

Common denominator!

 

[Hurkyl]

but I don't see how to manipulate (2) to make it equal to (1)

You don't need to. All you need to do is to show these two sums of fractions are equal.

 

[livcon]

That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks

 

[Hurkyl]

As a warmup exercise, prove the following:

\frac{1}{3} + \frac{1}{6} = \frac{3}{2} - 1

 

[emyt]

That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks

see how 1 - 1/n+1 looks first

 

[livcon]

I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.

 

[Hurkyl]

I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.

Isn't proving

1 - \frac{1}{(n+1)+1} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}​

exactly the same kind of problem, though? Why can't you do the same thing you did for the warmup problem?

 

[livcon]

I guess it's because in the warmup exercise I could easily add and subtract the fractions because it was all constants, but in the one involving the variable n I got stuck. So I would probably have an epiphany if someone would just write it out. This is just a matter of me not seeing the obvious steps in this particular problem.

 

[HallsofIvy]

What do you get when you change
1- \frac{1}{n+1}+ \frac{1}{(n+1)(n+2)}
to fractions having the same denominator?

That's what everyone has been telling you do to. Have you done it?