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Math Help - Induction problem

  1. #1
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    Induction problem

    Q: Prove the set of { S = n \epsilon <b>N</b> : \frac{1}{1^2} + \frac{1}{2^2} + ... + \frac{1}{n^2} <= 2 - \frac{1}{n}} is inductive.

    P.S. Are there any website or program in which I can learn more about the math coding in here? I had been spending hours inputing simply math operations. For example, I had just typed out my progress of work so far in this problem, took me about half an hour, but after I hit save, it just disappeared, did not update it, not anything. So I'm not going to type out my progress, sorry!

    Thank you.
    Last edited by tttcomrader; November 18th 2006 at 09:29 AM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Q: Prove the set of { S = n \epsilon <b>N</b> : \frac{1}{1^2} + \frac{1}{2^2} + ... + \frac{1}{n^2} <= 2 - \frac{1}{n}} is inductive....
    Hello,

    If I understand your problem right, you want to prove this property by mathematical induction(?):

    I assume that you know how to do MI:

    1. Prove for n = 1: \frac{1}{1^2}=2-\frac{1}{1} is true.

    2. Assume that \frac{1}{1^2} + \frac{1}{2^2} + ... + \frac{1}{n^2} <= 2 - \frac{1}{n} is true.

    3. Prove that the step from n to (n+1) is true under the assumption:

    \frac{1}{1^2} + \frac{1}{2^2} + ... + \frac{1}{n^2}+ \frac{1}{(n+1)^2} <= 2 - \frac{1}{n}+ \frac{1}{(n+1)^2}

    <= 2 - \frac{(n+1)^2-n}{n \cdot (n+1)^2}. Now we add \frac{1}{n \cdot (n+1)^2} which is possible because of the <= sign!

    <= 2 - \frac{(n+1)^2-n-1}{n \cdot (n+1)^2}

    <= 2 - \frac{(n+1)^2-(n+1)}{n \cdot (n+1)^2}

    <= 2 - \frac{(n+1)((n+1)-1)}{n \cdot (n+1)^2}

    <= 2 - \frac{(n+1)n}{n \cdot (n+1)^2} . Cancel out the equal factors:

    <= 2 - \frac{1}{n+1}

    Thus this property is true for all n \in \mathbb{N}

    EB
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