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Proof Question: Using Mathematical Induction

  1. Mar 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that, for all integers [itex]n =>1[/itex]

    [tex]\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1} = 1-\frac{1}{n+1}[/tex]


    2. Relevant equations

    I am a little stuck on this question. :|

    3. The attempt at a solution
     
    Last edited: Mar 24, 2008
  2. jcsd
  3. Mar 24, 2008 #2

    Gib Z

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    Well, the LHS can be expressed as

    [tex]\sum_{k=1}^n \frac{1}{k(k+1)}[/tex]. Use partial fractions on that expression, you should get a telescoping series.
     
  4. Mar 25, 2008 #3
    Anyone got any ideas with this problem?

    I cant find any notes regarding a similar question.
     
  5. Mar 25, 2008 #4

    HallsofIvy

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    What was wrong with Gib Z's suggestion?
     
  6. Mar 25, 2008 #5
    We haven't been taught that.

    Im a little unsure what he means as well.

    Could you possibly give more help?

    Sorry if im a pain.
     
  7. Mar 25, 2008 #6
    Putting it in partial fractions gives.

    [itex]\frac{1}{k(k+1)}=\frac{1}{k}\cdot\frac{1}{(k+1)}[/itex] or [itex]\equiv[/itex]?

    Do you mean you don't understand summation?

    All that expression is is your first expression for any given value of n. It's essentially the same thing.
     
    Last edited: Mar 25, 2008
  8. Mar 25, 2008 #7

    tiny-tim

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    … give it a name … !

    Hi Lococard! :smile:

    For induction proofs, it often helps to give a name to the sum (if the question hasn't already doe so).

    In this case, use the name a_n+1:

    [tex]a_{n+1}\,=\,\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1}[/tex]

    Then what you have to prove is that, assuming that:
    a_n = 1 - 1/n,
    then:
    a_n+1 = 1 - 1/(n+1).​

    You see how that makes the problem easier? :smile:
     
  9. Mar 25, 2008 #8
    How would i put it into a partial fraction?
     
  10. Mar 25, 2008 #9
    Well I gave one partial fraction that is in fact equivalent to your original equation. Put that equation into your calculator, plug in some numbers and then use the original equation, if they are equivalent it follows that both equations are different versions of the same partial fraction. Of course you could do this algebraically too, but I'll leave that to you. Tiny-tim has given you a nice hint there.
     
  11. Mar 25, 2008 #10
    I've done them both on paper and i got the same result.

    Do i then add a_n+1 to the equation and get the formula in the lowest form?
     
  12. Mar 26, 2008 #11

    tiny-tim

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    Hi Lococard! :smile:

    I'm not sure what you mean. :confused:

    Just say a_n+1 - a_n = 1/n(n+1);

    then, from the induction assumption, a_n = 1 - 1/n = (n - 1)/n.

    So a_n+1 = a_n + 1/n(n+1) = (n - 1)/n + 1/n(n+1) = (n^2 - 1)/n(n+1) + 1/n(n+1) = … ? :smile:

    Is that what you meant?
     
  13. Mar 26, 2008 #12
    Alright, i got some more info on the question.

    It seems that:

    [tex]\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1} = 1-\frac{1}{n+1}[/tex]


    But. The number after [tex] \frac{1}{n+1}[/tex] would be [tex] \frac{(n+1)}{(n+1)(n+2)}[/tex]

    So.

    [tex]1 - \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)}[/tex]

    = 1 - ??

    Would would i simplify the rest of it?
     
  14. Mar 26, 2008 #13

    tiny-tim

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    ah! … you have a misprint …

    it should be [tex]\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n*(n+1)}\,=\,1-\frac{1}{n+1}\,.[/tex]

    Does that look better? :smile:
     
  15. Mar 26, 2008 #14
    Oh whoops. Yeah thats what i meant
     
  16. Mar 26, 2008 #15
    How would i do the next part?
     
  17. Mar 26, 2008 #16

    tiny-tim

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    [tex]1 - \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)}[/tex]

    = [tex] \frac{n+1}{(n+1)} - \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)}[/tex]

    = … ? :smile:

    (btw, wouldn't it have been easier to work out if you'd "gone down one", and writen:
    [tex]1 - \frac{1}{n} + \frac{1}{n(n+1)}[/tex] ?)
     
  18. Mar 26, 2008 #17
    did you just substitute the 1 for a [itex]\frac{n+1}{n+1}[/itex]?


    Im really stuck on simplifying the RHS.
     
  19. Mar 26, 2008 #18

    tiny-tim

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    Yes!

    Tell me what bothers you about that …

    Anyway, the next step is to use

    [tex] \frac{n+1}{(n+1)} - \frac{1}{(n+1)}\,=\,\frac{n}{(n+1)}\,.[/tex]

    And then … ? :smile:
     
  20. Mar 26, 2008 #19
    How did you do that step?

    Now im really lost.


    The lecturer suggested i get the equation to = [itex] 1 - \frac{1}{n+2}[/itex]
     
  21. Mar 26, 2008 #20

    tiny-tim

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    ah!

    You don't like polynomial fractions, do you?

    They're just like ordinary fractions …

    If you had 1 - 15/17, you'd say that's 17/17 - 15/17, = 2/17.

    Similarly, if you had 1 - 15/(n+17), you'd say that's (n+17)/(n+17) - 15/(n+17), = (n+17-15)/(n+17) = (n+2)/(n+17).

    You gotta practise this, and become happy with it!

    So … 1 - 1/(n+1) = … ? :smile:
     
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