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Thread: finding the Inequality of arithmetic and geometric means of a sequence

  1. #1
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    finding the Inequality of arithmetic and geometric means of a sequence

    Hello everyone


    I was requested to prove this inequality

    finding the Inequality of arithmetic and geometric means of a sequence-photo.png


    for this sequence


    finding the Inequality of arithmetic and geometric means of a sequence-photo2.png


    using Inequality arithmetic and geometric means (using arithmetic mean and geometric mean only)

    I tried to solve this thing out in so many ways I will be really glad if someone will show me the full solution for this Inequality


    thanks
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  2. #2
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    Re: finding the Inequality of arithmetic and geometric means of a sequence

    Quote Originally Posted by someone111888 View Post
    I was requested to prove this inequality
    Click image for larger version. 

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    for this sequence
    Click image for larger version. 

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ID:	38444
    As posted, the statement is not true.

    $a_1=\dfrac{1}{2},~a_2=\dfrac{7}{12},~a_3=\dfrac{3 7}{60},\cdots~a_9=\dfrac{1632341}{2450448}$

    Could it be that $N\ge 10~?$ Are you sure that you have copied it correctly?
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    Re: finding the Inequality of arithmetic and geometric means of a sequence

    Yes I did, my intesntion is that there is a requestment to prove that the sequence itself bigger than 10 using this equality
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    Re: finding the Inequality of arithmetic and geometric means of a sequence

    Im sorry bigger than 2/3
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    Re: finding the Inequality of arithmetic and geometric means of a sequence

    Quote Originally Posted by someone111888 View Post
    Yes I did, my intesntion is that there is a requestment to prove that the sequence itself bigger than 10 using this equality
    Quote Originally Posted by someone111888 View Post
    Im sorry bigger than 2/3
    Now I am totally confused.

    Just post the exact wording of the original question.
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  6. #6
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    Re: finding the Inequality of arithmetic and geometric means of a sequence

    Quote Originally Posted by Plato View Post
    Now I am totally confused.
    What else is new?
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  7. #7
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    Re: finding the Inequality of arithmetic and geometric means of a sequence

    Quote Originally Posted by someone111888 View Post
    Im sorry bigger than 2/3
    Then DenisB just gave you a proof by contradiction showing that inequality is false for n <10. You cannot prove a false statement to be true. He gave you explicit examples showing that it is false.


    $a_{n+1}=a_n+\dfrac{1}{2n+1}+\dfrac{1}{2n+2}-\dfrac{1}{n+1} = a_n+\dfrac{1}{(2n+1)(2n+2)} $

    So your sequence is increasing as n increases. You should be able to find the minimum value of n that makes the equality true

    http://m.wolframalpha.com/input/?i=l...to+infinity%5D
    Last edited by SlipEternal; Jan 3rd 2018 at 03:20 PM.
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  8. #8
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    Re: finding the Inequality of arithmetic and geometric means of a sequence

    Notice first that for any positive integer n, $\int_n^{2n}1/x\,dx=\ln(2)$. Next by consideration of upper and lower Riemann sums for this integral, it is easy to show that
    $$\ln(2)-{1\over2n}\leq a_n\leq\ln(2)$$

    Hence
    $$\lim_{n\to\infty}a_n=\ln(2)$$

    Since $\ln(2)\approx0.693$, from the above inequality, it is obvious that eventually $a_n>2/3$. As was noted in above posts, the inequality is true for any $n\geq10$. To me, this was kind of a strange question; I think a better question would be to find the limit of the sequence, as done above.
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