Inductive proof of summation formula

  • Thread starter livcon
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  • #1
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Homework Statement


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

Homework Equations


-

The Attempt at a Solution


Inductive step:
1. [tex]\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}[/tex]

2. [tex]\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)}[/tex]

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!
 

Answers and Replies

  • #2
whs
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Common denominator!
 
  • #3
Hurkyl
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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.
 
  • #4
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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
 
  • #5
Hurkyl
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As a warmup exercise, prove the following:

[tex]\frac{1}{3} + \frac{1}{6} = \frac{3}{2} - 1[/tex]
 
  • #6
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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
 
  • #7
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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.
 
  • #8
Hurkyl
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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
[tex]1 - \frac{1}{(n+1)+1} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}[/tex]​
exactly the same kind of problem, though? Why can't you do the same thing you did for the warmup problem?
 
  • #9
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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.
 
  • #10
HallsofIvy
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What do you get when you change
[tex]1- \frac{1}{n+1}+ \frac{1}{(n+1)(n+2)}[/tex]
to fractions having the same denominator?

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

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