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Inductive proof help on last step

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data

    i=1 Sigma n (1/i2) <= 2 - (1/n)


    3. The attempt at a solution

    I've done the basic step and assumption step...little stuck on the inductive step

    So far I have...

    show 1 + 1/4 + 1/9 + 1/16 +...+ (1/k2) + (1/(k+1)2) <= 2 - (1/k+1)
     
  2. jcsd
  3. Sep 3, 2012 #2

    SammyS

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    To clarify matters:

    I take it that you need to prove (by induction) that:
    [itex]\displaystyle
    \sum_{i=1}^{n}\frac{1}{i^2}\le2-\frac{1}{n}\ .[/itex]​

    Is that correct?

    So, you have assumed that 1 + 1/4 + 1/9 + 1/16 +...+ (1/k2) ≤ 2 - (1/k) ,

    and you need to show that 1 + 1/4 + 1/9 + 1/16 +...+ (1/k2) + (1/(k+1)2) ≤ 2 - (1/(k+1)) .

    Is that correct?

    What have you tried, in this effort?

    BTW: Please learn to use parentheses, so that your mathematical expressions say what you mean for them to say.
     
  4. Sep 4, 2012 #3
    Yes, that is correct...I've done equality inductive proofs, but have not encountered less than or greater than type proofs...so I'm not sure how to begin.
     
  5. Sep 4, 2012 #4
    The idea is almost exactly the same as equality. Starting from where you left off you have [itex]\sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{n+1}[/itex]≤ 2- [itex] \frac{1}{n+1}[/itex]
    See what happens if you combine like terms and use what you already know about [itex]\sum_{i=1}^{n}\frac{1}{i^2}[/itex] to help you out.
     
  6. Sep 4, 2012 #5

    SammyS

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    Let's see:

    You are assuming that some k,
    1 + 1/4 + 1/9 + 1/16 +...+ 1/k2 ≤ 2 - (1/k) .​

    Adding 1/(k+1)2 to both sides gives you that
    1 + 1/4 + 1/9 + 1/16 +...+ 1/k2 + 1/(k+1)2 ≤ 2 - (1/k) +1/(k+1)2 .​

    So, if you can show that 2 - (1/k) +1/(k+1)2 ≤ 2 - 1/(k+1), then you should be able to complete the proof.
     
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