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Loser66

  • one year ago

Use induction to prove \(\sum_{k=1}^n \dfrac{1}{k^2}\leq 2-\dfrac{1}{n}\) for all positive integers n. Please, help

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  1. Loser66
    • one year ago
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    HS: \(\sum_{k=1}^n \dfrac{1}{k^2}\leq 2 - \dfrac{1}{n}\) Induction step \(\sum_{k=1}^{n+1} \dfrac{1}{k^2}= \sum_{k=1}^n \dfrac{1}{k^2}+\dfrac{1}{(n+1)^2}\leq 2-\dfrac{1}{(n+1)^2}\)

  2. Loser66
    • one year ago
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    oh, the last one is not ^2 \(\sum_{k=1}^{n+1} \dfrac{1}{k^2}= \sum_{k=1}^n \dfrac{1}{k^2}+\dfrac{1}{(n+1)^2}\leq 2-\dfrac{1}{(n+1)}\)

  3. Loser66
    • one year ago
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    This is my attempt: the first part of the sum is \(\leq 2 - (1/n)\) , hence the LHS is \(\leq 2 -(1/n) + (1/n+1)^2\)

  4. Loser66
    • one year ago
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    So, we need prove \(2-\dfrac{1}{n} +\dfrac{1}{(n+1)^2}\leq 2- \dfrac{1}{n+1}\)

  5. Loser66
    • one year ago
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    Then I stuck

  6. ganeshie8
    • one year ago
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    clear the denominators and simplify

  7. ganeshie8
    • one year ago
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    it simplifies nicely and you end up with \(1\ge 0\)

  8. Loser66
    • one year ago
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    Yes, I did, but those are denominators, hence when I take reciprocal, the signs switch around.

  9. Loser66
    • one year ago
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    oh, let me try again. :)

  10. ganeshie8
    • one year ago
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    \(2-\dfrac{1}{n} +\dfrac{1}{(n+1)^2}\leq 2- \dfrac{1}{n+1}\) \(\iff\) \(\dfrac{1}{n} -\dfrac{1}{(n+1)^2}\ge \dfrac{1}{n+1}\) \(\iff\) \((n+1)^2-n\ge n(n+1)\)

  11. Loser66
    • one year ago
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    Thanks a lot.

  12. Loser66
    • one year ago
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    I got it. Thank you so much

  13. ganeshie8
    • one year ago
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    np

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