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Math Help - Help with induction problem (inequality)

  1. #1
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    Help with induction problem (inequality)

    Hi, problem is for any n >= 2, 1/n+1 + 1/n+2 + 1/n+3 + ... + 1/2n > 13/24 . I am having trouble putting into summation notation which is required first and then beginning. Help is appreciated, thanks
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  2. #2
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    Quote Originally Posted by kanoro View Post
    Hi, problem is for any n >= 2, 1/n+1 + 1/n+2 + 1/n+3 + ... + 1/2n > 13/24 . I am having trouble putting into summation notation which is required first and then beginning. Help is appreciated, thanks
    \displaystyle\sum_{m=1}^n\frac{1}{n+m}>\frac{13}{2  4},\;\;\;\;n\ \ge\ 2

    P(k)

    \displaystyle\sum_{m=1}^k\frac{1}{k+m}>\frac{13}{2  4}

    P(k+1)

    \displaystyle\sum_{m=1}^{k+1}\frac{1}{k+1+m}>\frac  {13}{24}

    Try to prove that P(k+1) will be valid if P(k) is and test for the first n.
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  3. #3
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    Might not be a big thing, but do not forget to show that it is true for a specific n e.g. n = 2.
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    I understand it all, but when I am proving the initial case for n = 2, what value will the m be to make the inequality true?
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  5. #5
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    Quote Originally Posted by kanoro View Post
    I understand it all, but when I am proving the initial case for n = 2, what value will the m be to make the inequality true?
    Your values of m go from 1 to n, so for n=2, you take the first 2 positive integers, m=1 and m=2.
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  6. #6
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    m is your index number for the sum. Thus m takes any value between 1 and n. The basic idea of Sigma ( \sum) is by example:

    \sum_{k=1}^{10} k = 1+2+3+4+5+6+7+8+9+10

    Just fill in values similarily-
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  7. #7
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    Quote Originally Posted by kanoro View Post
    I understand it all, but when I am proving the initial case for n = 2, what value will the m be to make the inequality true?
    I forgot to answer the rest of this....

    it isn't a matter of certain values of "m" making the inequality true.
    You simply sum all the fractions (increasing m by 1 in each fraction) until m reaches n (not necessary when proving by induction).

    For all n greater than or equal to 2, your sum ought to be greater than 13/24.
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  8. #8
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    I'm working on solving the inequality but I am a little stuck.. I have the 1 / k + 1 + m > 13/24 and I understand the operations that can be applied to inequalities while maintaining to hold their values. Any clues on how to approach this?
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  9. #9
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    Note that:
    \displaystyle\sum_{m=1}^{k+1}\frac{1}{k+(1+m)} = \displaystyle\sum_{m=2}^{k+2}\frac{1}{k+m} = \displaystyle\sum_{m=1}^{k}\frac{1}{k+m} + \frac{1}{k+k+2} + \frac{1}{k+k+1}- \frac{1}{k+1}

    Now you should be able to solve.
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  10. #10
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    Hmm, I am not seeing where this helps. Lost me there.
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  11. #11
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    Quote Originally Posted by kanoro View Post
    I'm working on solving the inequality but I am a little stuck.. I have the 1 / k + 1 + m > 13/24 and I understand the operations that can be applied to inequalities while maintaining to hold their values. Any clues on how to approach this?
    You break down the P(k+1) to

    \displaystyle\sum_{m=1}^{k+1}\frac{1}{k+1+m}=\sum_  {m=1}^k\frac{1}{k+1+m}+\frac{1}{k+1+k+1}

    =\displaystyle\sum_{m=1}^k\frac{1}{m+k}-\frac{1}{k+1}+\frac{1}{k+k+1}+\frac{1}{2k+2}

    which raphw has also shown and now you use P(k) to show that if P(k) is true,
    then P(k+1) will also be true.

    Add the three fractions on the right to show they sum to a value >0
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  12. #12
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    So then I get summation as m=1 goes to k, 1/m+k - 8k+5 / ( 2k+2)(2k+1) . So how do I prove that they sum >0
    Last edited by kanoro; October 11th 2010 at 07:36 PM.
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  13. #13
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    In general still confused where to take this, textbook I am working out offers little help so any help to finish the question is much appreciated.
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  14. #14
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    The general idea of induction is as follows: (1) I show that something is true for a specific value of n, e.g. n=2. Then (2) I assume that this same statement is true for any n and (3) I show that it is then true for n+1. It follows that it must be true for any value since I can start by e.g. n=2 => true for n+1 = 3 => true for n+2 = 4 => ...

    So we showed that it is true for n=2 already. Now we assume that it is true for \displaystyle\sum_{m=1}^{k}\frac{1}{k+m}

    Thus we can solve
    \displaystyle\sum_{m=1}^{k+1}\frac{1}{k+(1+m)} = \displaystyle\sum_{m=2}^{k+2}\frac{1}{k+m} = \displaystyle\sum_{m=1}^{k}\frac{1}{k+m} + \frac{1}{k+k+2} + \frac{1}{k+k+1}- \frac{1}{k+1} > \frac{13}{24} + \frac{1}{k+k+2} + \frac{1}{k+k+1}- \frac{1}{k+1} > \frac{13}{24}

    Thus:
    \displaystyle\sum_{m=1}^{k+1}\frac{1}{k+(1+m)} > \frac{13}{24}
    and we completed our proof by induction since we can now construct any n as shown above.

    This last step is true if and only if \frac{1}{k+k+2} + \frac{1}{k+k+1}- \frac{1}{k+1} \ge 0 This is all that is left to show to finish the proof. To do so, simply get a common denominator. If both numerator and denominator of the new, "merged" term are positive, you are done.
    Last edited by raphw; October 11th 2010 at 08:13 PM.
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  15. #15
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    Quote Originally Posted by raphw View Post
    The general idea of induction is as follows: (1) I show that something is true for a specific value of n, e.g. n=2. Then (2) I assume that this same statement is true for any n and (3) I show that it is then true for n+1. It follows that it must be true for any value since I can start by e.g. n=2 => true for n+1 = 3 => true for n+2 = 4 => ...
    I know that this is semantics and you probably do understand that what you wrote is wrong - but what you're actually doing is proving that if the statement is true for some n \ge 2, then it is true for n+1.
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